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#### Page No 76:

No, a gas doesn't have just two specific heat capacities, as the heat capacities depend on the process followed. There are infinite processes; therefore, there can be infinite number of specific heat capacities.

#### Page No 76:

Specific heat capacity, , where $\Delta$Q/m is the heat supplied per unit mass of the substance and $\mathrm{\Delta }$T is the change in temperature produced. At constant temperature, ​$\mathrm{\Delta }$T = 0; therefore, s = infinity. So, we cannot define specific heat capacity at constant temperature.

#### Page No 76:

Specific heat capacity, , where$∆$Q/m is the heat supplied per unit mass of the substance and $∆$T is the change in temperature produced. In an adiabatic process, no heat exchange is allowed; so, $∆$Q = 0 and hence, s = 0. Therefore, in an adiabatic process, specific heat capacity is zero.

#### Page No 76:

Yes, a solid also has two kinds of molar heat capacities, Cp and Cv. In a solid, expansion coefficient is quite small; therefore dependence of heat capacity on the process is negligible. So, Cp Cv with just a small difference, which is not equal to R.

#### Page No 76:

In a real gas, as the internal energy depends on temperature and volume, the derived equation for an ideal gas  will change to ​ + k, where k is the change in internal energy (positive) due to change in volume when pressure is kept constant. So, in the case of a real gas, for n=1 mole (say),

where Cp and Cv are the specific heat capacities at constant pressure and volume, respectively.

#### Page No 76:

According to the first law of thermodynamics, change in internal energy, $∆$U is equal to the difference between heat supplied to the gas, $∆$Q and the work done on the gas,​$∆$W, such that $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$​. In an adiabatic process, $∆$Q = 0 and in an isothermal process, change in temperature, $∆$T = 0. Therefore,
$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta }Q=n{C}_{v}\mathrm{\Delta }T+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒0=n{C}_{v}\left(0\right)+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta }W=0,$
where Cv is the heat capacity at constant volume.
This shows that if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.

#### Page No 76:

In an isothermal process,
PV = k
...(i)
On differentiating it w.r.t V, we get

k = constant

On differentiating it w.r.t V, we get

K is constant
are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively; P  is pressure and V is volume of the system.
By comparing the two slopes and keeping in mind that $\gamma >1$, we can see that the slope of the P-V diagram is greater for an adiabatic process than an isothermal process.

#### Page No 76:

For an isothermal process, PV = K, where P is pressure, V is volume of the system and K is constant. In an isothermal process, a small change in V produces only a small change in p, so as to keep the product constant. On the other hand, in an adiabatic process,  > 1 is the ratio of  heat capacities at constant pressure and volume, respectively, and k is a constant. In this process, a small increase in volume produces a large decrease in pressure. Therefore, an isothermal process is considered to be a slow process and an adiabatic process a quick process.

#### Page No 76:

For two states to be connected by an isothermal process,
${P}_{1}{V}_{1}={P}_{2}{V}_{2}$   ...(i)
For the same two states to be connected by an adiabatic process,
...(ii)
If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get

Let the gas be monatomic. Then,

If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.

#### Page No 76:

For the molecules of a gas, $\gamma =\frac{{C}_{\mathrm{p}}}{{C}_{\mathrm{v}}}=1+\frac{2}{f}$,
where f is the degree of freedom.

Therefore, the molecules of this gas have 7 degrees of freedom.
But in reality, no gas can have more than 6 degrees of freedom.

#### Page No 76:

(c) C> CB

According to the first law of thermodynamics, $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$, where $∆$Q is the heat supplied to the system when $\Delta$W work is done on the system and $∆$U is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. $∆$UA = $∆$UB.
But as, $∆$WA$∆$WB this gives $\Delta$QA = 2$\Delta$QB.
Now, molar heat capacity of a gas, , where $∆$Q/n is the heat supplied to a mole of gas and $∆$T is the change in temperature produced. As $∆$QA = 2$∆$QB, C> CB.

#### Page No 76:

c) Cp is slightly greater than Cv

For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore, Cp is slightly greater than Cv.

#### Page No 76:

(a) pA < pB and TA > TB

CpC= R for the gas in state A, which means it is acting as an ideal gas in that state, whereas Cp − Cv = 1.08in state B, i.e. the behaviour of the gas is that of a real gas in that state. To be an ideal gas, a real gas at STP should be at a very high temperature and low pressure. Therefore, PA < PB and TA > TB where PA and PB denotes the pressure and TA and TB denotes the temperature of system A and B reepectively.

#### Page No 76:

(c) Cp Cv

For an ideal gas, Cp Cv = R , where Cv and Cp denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively  and R is the gas constant whos value is 8.314 J/K. Therefore, Cp Cv is a constant. On the other hand, the ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.

#### Page No 76:

(b) 50 calories
It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure,

For an ideal gas,

Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35 oC at constant volume.

#### Page No 76: (c) C > C​v

Consider two processes AB and ACB; let W be the work done. C is the molar heat capacity of process AB. Process ACB can be considered as the sum of the two processes, AC and CB. The molar heat capacity of process AC is Cp, as pressure is constant in this process and the molar heat capacity of process CB is Cv, as volume is constant in it.
Internal energy, U, is a state function, i.e. it doesn't depend on the path followed. Therefore,

Work done in the p-V diagram is the area enclosed under the curve.
$⇒{W}_{\mathrm{AB}}+{U}_{\mathrm{AB}}>{W}_{\mathrm{ACB}}+{U}_{\mathrm{ACB}}\phantom{\rule{0ex}{0ex}}⇒C>{C}_{\mathrm{v}}+{C}_{\mathrm{p}}$
Molar heat capacity is the heat supplied per mole to change the temperature by a degree Kelvin and according to the first law of thermodynamics, dQ = dU + dW, where dQ is the heat supplied to the system in a process.
$⇒C>{C}_{\mathrm{v}}$

#### Page No 76:

(d) C = 0.

The defined process is
$p=\frac{k}{{V}^{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}⇒p{V}^{\mathrm{g}}=k,$
such that the process is adiabatic in which there's no heat supplied to the system, i.e. Q = 0. Molar heat capacity is the amount of heat supplied to the system per mole to produce a degree change in temperature. Also, in an adiabatic process, no heat exchange is allowed. So, molar heat capacity equals zero, i.e. C = 0.

#### Page No 76:

(b) 0.5%

Let p and p' be the initial and final pressures of the system and V and V' be the initial and final volumes of the system.  p' is 0.5% more than p and the process is isothernal. So, pV = k = p'V' =  constant. Therefore,

So, volume V' decreases by about 0.5% of V.

#### Page No 76:

(a) 0.36 %

Let p and p, be the initial and final pressures of the system and V and V, be the initial and final volumes of the system. p, is 0.5% more than p and the process is adiabatic. So,

Therefore, V, is 0.36 % less than V.

#### Page No 76:

(c) pA < pB Let the initial states of samples A and B be i and the final states of samples B and A be f and f', respectively. Let the final volumes of both be Vo. As sample A is expanded through an adiabatic process, its curve in the p-V diagram is steeper than that of sample B, which is expanded through an isothermal process. Therefore, from the p-V diagram, pA < pB.

#### Page No 76:

(a) Ta < Tb
As sample B is undergoing expansion through an isothermal process, its initial and final temperatures will be same, i.e. Tb. On the other hand, sample A is at the same initial state as B, such that the initial temperature of A is Tb and it is expanding through an adiabatic process in which no heat is supplied. Therefore, sample A will expand at the cost of its internal energy and its final temperature will be less than its initial temperature.
This implies that  Ta < Tb.

#### Page No 76:

(c) ∆Wa < ∆Wb In the p-V diagram, the area under the curve w.r.t the V axis is equal to the work done by the system. Since the area under the isotherm is greater than that under the adiabat, the work done by system A is less than that done by system B. Hence,  ∆Wa < ∆Wb.

#### Page No 77:

(d) molecular vibrations gradually become effective

Molar specific heat capacity has direct dependence on the degree of freedom of gas molecules. As temperature is increased, the gas molecules start vibrating about their mean position, leading to change (increase) in the degree of freedom and, hence, increasing molar heat capacity.

#### Page No 77:

(c) may be very nearly adiabatic
(d) may be very nearly isothermal
Due to sudden compression, the gas did not get sufficient time for heat exchange. So, no heat exchange occurred. Therefore, the process may be adiabatic. For any process to be isothermal, its temperature should remain constant, i.e. pressure and volume should change simultaneously while their product (temperature) should be constant.

#### Page No 77:

(d) Q = W

In an isothermal process, temperature of the system stays constant, i.e. there's no change in internal energy. Thus, U = 0, where U denotes the change in internal energy of the system. According to the first law of thermodynamics, heat supplied to the system is equal to the sum of change in internal energy and work done by the system, such that Q = U + W. As U = 0, Q = W.

#### Page No 77:

(a) Q = 0
(d) QW
In an adiabatic process, no heat is supplied to the system; so, Q = 0. According to the first law of thermodynamics, heat given to any system is equal to the sum of the change in internal energy and the work done on the system. So, Q = W+U and as Q = 0, W = -and  QW.

#### Page No 77:

(c) A is isothermal and B is adiabatic
The slope of an adiabatic process is greater than that of an isothermal process. Since A and B are initiated from the same initial state, both cannot be isothermal or adiabatic, as they would be overlapping. But the curve of process B is steeper than the curve of process A. Hence, A is isothermal and B is adiabatic.

#### Page No 77:

(c) The pressures of helium and neon will be the same but that of oxygen will be different.
(d) The temperatures of helium and neon will be the same but that of oxygen will be different.

where $\gamma =\frac{{C}_{p}}{{C}_{v}}$ is the ratio of molar heat capacities at constant pressure and volume.
We know that $\gamma$ is equal to 1.67 and 1.40 for a monatomic gas and a diatomic gas, respectively. Helium and neon are monatomic gases and oxygen is a diatomic gas. Therefore, changing the state of the gases, i.e. reducing the volume will lead to identical changes in temperature and pressure for helium and neon and that will be different for oxygen.

#### Page No 77:

(a) helium
(b) argon

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54 J of heat is added to it. So, its molar heat capacity, C​v = 12.54 J $J{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas, ​C​v $\simeq$= 12.54 JK-1mol-1. Among the given gases, only helium and argon are inert and, hence, monoatomic. Therefore, the gas may be helium or argon.

#### Page No 77:

(a) argon

The energy of a gas is measured as CvTAll the four cylinders are at the same temperature but the gases in them have different values of  Cv, such that it is least for the monatomic gas and keeps on increasing as we go from monatomic to tri-atomic. Among the above gases, argon is monatomic, hydrogen and nitrogen are diatomic and carbon dioxide is tri-atomic. Therefore, the energy is minimum in argon.

#### Page No 77:

Number of moles of the ideal gas, n = 1 mole
Molecular weight of the gas, W = 20 g/mole
Mass of the gas, m =20 g
Velocity of the vessel, V = 50 m/s
Decrease in K.E. of the vessel = Internal energy gained by the gas

#### Page No 77:

Given:
Mass of the gas, m = 5 g
Change in temperature of the system, ∆T = 25 − 15°C = 10°C
Specific heat at constant volume, Cv = 0.172 cal/g -°C
Mechanical equivalent, J = 4.2 J/cal

From the first law of thermodynamics,
dQ =  dU + dW
Now,
$∆$V = 0 (Rigid wall of the container keeps the volume constant)
So, dW = P$∆V$= 0
Therefore,
dQ = dU  (From the first law)
Q = m${c}_{v}$dT = 5 × 0.172 × 10
= 8.6 cal = 8.6 × 4.2 J
= 36.12 J
So, change in internal energy of the system is 36.12 J.

#### Page No 77:

Given:
Mass of the piston (m) = 50 kg
Adiabatic constant of the gas, γ = 1.4
Area of cross-section of the piston (A) = 100 cm2
Atmospheric pressure (P0) = 100 kPa
g = 10 m/s2
Distance moved by the piston , x = 20 cm
Work done by the gas,

W = (5 × 104 + 105) × 20 × 10−4
W = 1.5 × 105 × 20 × 10−4
W = 300 J

Hence, nRdT = P$∆V$ = 300

#### Page No 77:

Specific heat capacity at constant volume, Cv(H2) = 2.4 cal/g-°C
Specific heat capacity at constant pressure, Cp(H2) = 3.4 cal/g-°C
Molecular weight, M = 2 g/mol
Gas constant, R = 8.3 × 107 erg/mol-°C
We know: Cp − Cv = 1 cal/g-°C,
where Cp and Cv are molar specific heat capacities.
So, difference of molar specific heat,
Cp × M − Cv × M = 1 cal/mol-°C

Now, 2 × J = R
⇒ 2 × J = 8.3 × 107 erg/mol-°C
J = 4.15 × 107 erg/cal

#### Page No 77:

Given:
$\left(\frac{{\mathrm{C}}_{p}}{{\mathrm{C}}_{v}}\right)=\frac{7}{6}$
Number of moles of the gas, n = 1 mol
Change in temperature of the gas, ∆T = 50 K

(a) Keeping the pressure constant:
Using the first law of thermodynamics,
dQ = dU + dW
T = 50 K and $\gamma =\frac{7}{6}$
dQ = dU + dW
Work done, dW = PdV
As pressure is kept constant, work done = P($\mathit{∆}V$)
Using the ideal gas equation PV = nRT,
P($\mathit{∆}V$) = nR($∆$T)
dW = nR($∆$T)
At constant pressure,  dQ = nCp dT
Substituting these values in the first law of thermodynamics, we get
nCp dT = dU + RdT
dU = nCp dT − RdT
Using
$dU=1×\frac{R\gamma }{\left(\gamma -1\right)}×d\mathrm{T}-Rd\mathrm{T}$
= 7 RdT − RdT
= 7 RdT − RdT = 6 RdT
= 6 × 8.3 × 50 = 2490 J

(b) Keeping volume constant:
Work done = 0
Using the first law of thermodynamics,
dU  = dQ
dU = nCv dT

$=1×\frac{R}{\gamma -1}×d\mathrm{T}\phantom{\rule{0ex}{0ex}}=1×\left(\frac{8.3}{\frac{7}{6}-1}\right)×50$
= 8.3 × 50 × 6 = 2490 J

Using the first law of thermodynamics, we get
dU = − dW
$=\left[\frac{n×R}{\gamma -1}\left({T}_{1}-{T}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1×8.3}{7/6-1}=\left({T}_{2}-{T}_{1}\right)$
= 8.3 × 6 × 50 = 2490 J

#### Page No 78:

Here,

m = 1.18 g = 1.18×10-3 kg

ΔQ = 2.0×4.2 J

P = 1.0×106 Pa

V = 1.0×103 cm3 = 1.0×10-3 m3

T = 300K

Applying eqn. of state

PV = nRT

=> n = PV/RT

=> n = 1.0×105×1.0×10-3/(8.314×300)

=> n = 0.04

ΔT = 10C

(ΔH)v = nCv ΔT

2.0×4.2 = nCv×1

=> Cv = 8.4/n= 8.4/0.04

=> Cv = 210

Again we know

Cp – Cv =R

=> Cp = R + Cv

=> Cp = 8.3 + 210

=> Cp = 218.3

Now at constant pressure

(ΔH)p = nCp ΔT

=> (ΔH)p = 218.3×0.04×1 = 8.732 J

In calories

=> (ΔH)p = 8.732/4.2 = 2.08 cal

#### Page No 78:

Initial volume of the gas, V1 = 100 cm3
Final volume = V2 = 200 cm3
Pressure = 2 × 105 Pa
Heat supplied, dQ = 50 J

(a) According to the first law of thermodynamics,
dQ = dU + dW
dW = P$∆V=2×{10}^{5}×\left(200-100\right)×{10}^{-6}=20$
⇒ 50 = dU + 2 × 10
dU = 30 J

(b) For a monatomic gas,
U = $\frac{3}{2}n$RT
30 = n × $\frac{3}{2}$ × 8.3 × 300

$⇒n=\frac{2}{\left(83×3\right)}=\frac{2}{249}=0.008$

(c)  Also,
dU = nCvdT
$⇒{C}_{\mathit{v}}\mathit{=}\frac{\mathit{d}\mathit{U}}{\mathit{n}\mathit{d}\mathit{T}}=\frac{30}{0.008×300}=12.5$
Cp = Cv + R = 12.5 + 8.3 = 20.8 J/mol-K

(d) Cv = 12.5 J/mol-K

#### Page No 78:

Given:
Amount of heat given to the gas = Q
So, ∆Q = Q
Work done by the gas, ∆W = $\frac{\mathrm{Q}}{2}$
From the first law of thermodynamics,
Q = ∆W + ∆U
$⇒∆U=\mathrm{Q}-\frac{\mathrm{Q}}{2}=\frac{\mathrm{Q}}{2}$
For a monoatomic gas,
U = $\frac{3}{2}$ nRdT
$⇒\frac{\mathrm{Q}}{2}=ndT×\frac{3}{2}R$
$⇒$Q = 3nRdT

Again, for expansion at constant pressure,
Q = nCpdT,
where Cp is the molar heat capacity at constant pressure.
So, 3RndT = nCpdT
Cp = 3R

#### Page No 78:

Relation between pressure and volume of a gas is P = kV.

Ideal gas equation is PV = nRT.

For simplicity, take the number of moles of  a gas, n = 1.
RdT = 2 kVdV
$⇒\frac{RdT}{2kV}=dV$

From the first law of thermodynamics,
dQ = dU + dW
n${C}_{p}$dT = Cv dT + PdV

#### Page No 78:

As the process has specific heat capacity zero, the process is essentially an adiabatic process. #### Page No 78:

For the first ideal gas,
Cp1 = specific heat at constant pressure
Cv1 = specific heat at constant volume
n1 = number of moles of the gas
$\frac{{C}_{{\mathrm{p}}_{1}}}{{C}_{{\mathrm{V}}_{1}}}$ = γ and Cp1 − Cv1 = R
γCv1Cv1 = R
Cv1 (γ − 1) = R
$⇒{\mathrm{Cv}}_{1}=\frac{R}{\left(\mathrm{\gamma }-1\right)}\phantom{\rule{0ex}{0ex}}C{\mathrm{p}}_{1}=\mathrm{\gamma }\frac{R}{\left(\mathrm{\gamma }-1\right)}$

For the second ideal gas,
Cp2 = specific heat at constant pressure
Cv2 = specific heat at constant volume
n2 = number of moles of the gas
$\frac{{\mathrm{C}}_{{\mathrm{p}}_{2}}}{{\mathrm{C}}_{{\mathrm{v}}_{2}}}$ = γ and Cp2Cv2 = R
γCv2Cv2 = R
Cv2 (γ − 1) = R

Given:
n1 = n2 = 1 : 2
dU1 = nCv1dt
dU2= 2nCv2dT

When the gases are mixed,
nCv1dT + 2nCv2dT = 3nCvdT
${\mathrm{C}}_{\mathrm{v}}=\frac{{\mathrm{C}}_{{v}_{1}}+2{\mathrm{C}}_{{v}_{2}}}{3}\phantom{\rule{0ex}{0ex}}=\frac{\frac{R}{\left(\mathrm{\gamma }-1\right)}+\frac{2R}{\left(\mathrm{\gamma }-1\right)}}{3}\phantom{\rule{0ex}{0ex}}=\frac{3R}{\left(\mathrm{\gamma }-1\right)3}=\frac{R}{\mathrm{\gamma }-1}$

Hence, Cp / Cv in the mixture is γ.

#### Page No 78:

Specific heat at constant pressure of helium, Cp' = 2.5 R
Specific heat at constant pressure of hydrogen, Cp" = 3.5 R
Specific heat at constant volume of helium, Cv' = 1.5 R
Specific heat at constant volume of hydrogen, Cv" = 2.5 R
n1 = n2 = 1 mol
$dU\mathit{=}n{C}_{v}dT$

For the mixture of two gases,

[n1 + n2 ]  CvdT = n1C'vdT + n2C"vdT,
where ${C}_{V}$ is the heat capacity of the mixture.
$⇒{C}_{v}\mathit{=}\frac{{n}_{1}C{\mathit{\text{'}}}_{v}\mathit{+}{n}_{2}C{\mathit{"}}_{v}}{{n}_{1}\mathit{+}{n}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1.5R+2.5R}{2}=2R$
Cp = Cv + R = 2R + R = 3R
$\mathrm{\gamma }=\frac{\mathrm{C}p}{\mathrm{C}v}=\frac{3R}{2R}=1.5$

#### Page No 78:

Given:
Number of moles of the gas, (a) Temperature at a = ${T}_{a}$
${P}_{\mathit{a}}{V}_{\mathit{a}}\mathit{=}nR{T}_{\mathit{a}}$

Similarly, temperature at b,

Similarly, temperature at c  is 480 K and at d  is 240 K.

(b) For process ab,
dQ = ncpdT
[Since ab is isobaric]

For line bc, volume is constant. So, it is an isochoric process.
dQ = dU + dW
[dW = 0, isochoric process]
dQ = dU = nCvdT
dQ= nCv (${T}_{c}-{T}_{b}\right)$

(c) Heat liberated in cd (isobaric process),
dQ = − nCpdT

Heat liberated in da (isochoric process),
$⇒$dQ = dU
Q= −nCvdT

#### Page No 78:

(a) For line ab, volume is constant.
So, from the ideal gas equation,
$\frac{{P}_{\mathit{1}}}{{T}_{\mathit{1}}}\mathit{=}\frac{{P}_{\mathit{2}}}{{T}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{100}{300}=\frac{200}{{\mathrm{T}}_{2}}\phantom{\rule{0ex}{0ex}}$ For line bc, pressure is constant.

(b)
As process ab is isochoric, ${W}_{ab}=0$.
During process bc,
P = 200 kPa

The volume is changing from 100 to 150 cm3 .
Therefore, work done = 50 × 10−6 × 200 × 103 J
= 10 J

(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nCvdT
$⇒$Heat supplied = nCvdT
Now,

(d) dQ = dU + dW
Now,
dU = dQdW
= Heat supplied − Work done
= (24.925 + 14.925) − 10
= 39.850 − 10 = 29.850 J.

#### Page No 78:

For Joly's differential steam calorimeter,
,
where
m2 = mass of steam condensed
m2 = 0.095 g
Latent heat of vapourization, L = 540 cal/g = 540 × 4.2 J/g
m1 = mass of gas present
m1 = 3 g
Initial temperature, θ1 = 20°C
Final temperature, θ2 = 100°C
$⇒{C}_{v}=\frac{0.095×540×4.2}{3×\left(100-20\right)}$
= 0.89 = 0.9 J/g-K

#### Page No 78:

Given,
γ = 1.5
Since the process is adiabatic, PVγ = constant.

(a) P1V1γ = P2V2γ
Given, V1 = 4 L
V2 = 3 L
We need to find $\frac{{P}_{2}}{{P}_{1}}$.
$⇒\frac{{\mathit{P}}_{\mathit{2}}}{{\mathit{P}}_{\mathit{1}}}\mathit{=}{\left(\frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{V}}_{\mathit{2}}}\right)}^{\mathit{\gamma }}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{1.5}=1.5396=1.54$

(b) Also, for an adiabatic process,
TVγ−1 = constant
T1V1γ−1 = T2V2γ−1
$⇒\frac{{T}_{\mathit{2}}}{{T}_{\mathit{1}}}\mathit{=}{\left(\frac{{V}_{\mathit{1}}}{{V}_{\mathit{2}}}\right)}^{\gamma \mathit{-}\mathit{1}}={\left(\frac{4}{3}\right)}^{0.5}=1.154$

#### Page No 78:

Initial pressure of the gas, P1 = 2.5 × 105 Pa
Initial temperature, T1 = 300 K
Initial volume, V1 = 100 cc

P1V1γ = P2V2γ
⇒ 2.5 × 105 × V1.5${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × P2
P2 = 7.07 × 105
= 7.1 × 105 Pa

(b) Also, for an adiabatic process,
T1V1γ−1 = T2V2γ1
⇒ 300 × (100)1.5−1 = T2 × ${\left(\frac{100}{2}\right)}^{1.5-1}$
= T2 × (50)1.5−1
⇒ 300 × 10 = T2   × 7.07
T2 = 424.32 K = 424 K

(c) Work done by the gas in the process,

#### Page No 78:

Given:
For air, γ = 1.4
Initial temperature of air, T1 = 20°C = 293 K
Initial pressure, P1 = 2 atm
Final pressure, P2 = 1 atm
The bursting of the tyre is an adiabatic process. For an adiabatic process,

P11γ × T11γ = P1γ × T2γ
(2)1−1.4 × (293)1.4 = (1)1−1.4 × T21.4
⇒ (2)−0.4 × (293)1.4 = T21.4
⇒ 2153.78 = T21.4
T2 = (2153.78)1/1.4
= 240.3 K

#### Page No 78:

Initial pressure of the gas, P1 = 100 kPa
Initial volume of the gas,V1 = 400 cm3
= 400 × 10−6 m3
Initial temperature of the gas, T1 = 300 K
$\gamma =\frac{\mathrm{C}p}{\mathrm{C}v}=1.5$

(a) The gas is suddenly compressed to volume, V2 = 100 cm3 .
So, this is an adiabatic process.
P1V1γ = P2V2γ
⇒ 105 × (400)1.5 = P2 (100)1.5
P2 = 105(4)1.5 = 800 kPa

Also,
T1Vγ−1 = T2V2γ−1
⇒ 300 × (400)1.5−1 = T2 (100)1.5−1
⇒ 300 × (400)0.5 = T2 (100)0.5
T2 = 600 K

(b) If the container is slowly compressed, the heat transfer is zero, even thought the walls are adiabatic.
Thus, the values remain same. Thus,
P2 = 800 kPa
T2 = 600 K

#### Page No 78:

Given:
For the gas, $\frac{\mathit{C}\mathit{p}}{\mathit{C}\mathit{v}}\mathit{=}\gamma$
Initial pressure of the gas = P0
Initial volume of the gas = V0

(a)
(i) As the gas is slowly compressed, its temperature will remain constant.
For isothermal compression,
P1V1 = P2V2
So, P0V0 = P2 $\frac{{\mathrm{V}}_{0}}{2}$P2 = 2P0

(ii) Sudden compression means that the gas could not get sufficient time to exchange heat with its surroundings. So, it is an adiabatiac compression.
P1V1γ = P2V2γ

= 2γ × 2P0 = P02γ+1

(b)
P1V1γ = P2V2γ
P0V0γ = P'${\left(\frac{{V}_{0}}{2}\right)}^{\gamma }$
P' = P02γ

(ii) Isothermal compression:
P1V1 = P2V2
${2}^{\gamma }{P}_{0}×\frac{{V}_{0}}{2}=P"\left(\frac{{V}_{0}}{2}\right)$
P" = P02γ × 2
⇒ P" = P02γ+1

#### Page No 78:

(a) Given,
Initial pressure of the gas = p0
Initial volume of the gas = ${V}_{0}$
For an isothermal process,
PV = constant
So, P1V1 = P2V2
${P}_{\mathit{2}}\mathit{=}\frac{{\mathit{P}}_{\mathit{0}}{\mathit{V}}_{\mathit{0}}}{\mathit{\left(}\frac{{P}_{0}}{2}\mathit{\right)}}=2{\mathrm{V}}_{0}$
For an adiabatic process, P3 = $\frac{{P}_{0}}{4}$, V3 = ?
P2V2γ = P3V3γ

(b) P1V1γ = P2V2γ

Again, for an isothermal process,
P2V2 = P3V3

#### Page No 78:

The ideal gas equation is
PV = nRT
Given, ${P}_{1}$ = 150 kPa = 150 × 103 Pa
${V}_{1}$ = 150 cm3 = 150 × 10−6 m3
${T}_{1}$= 300 K

(a)

(b)

(c) Given,
P1 = 150 kPa = 150 × 103 Pa
P2 = ? V1 = 150 cm3
= 150 × 10−6 m3
γ = 1.5
V2 = 50 cm3 = 50 × 10−6 m3,
T1 = 300 K
T2 = ?
Since the process is adiabatic, using the equation of an adiabatic process,we get
P1V1γ = P2V2γ
⇒ 150 × 103 × (150 × 10−6)γ = P2 × (50 × 10−6)γ
$⇒{P}_{2}=150×{10}^{3}×\frac{\left(150×{10}^{-6}{\right)}^{1.5}}{\left(50×{10}^{-6}{\right)}^{1.5}}$
P2 = 150000 × (3)1.5
P2 = 779.422 × 103 Pa
P2 = 780 kPa
Again,
P11−γ T1γ = P11−γ T2γ
⇒ (150 × 103)1−1.5 × (330)1.5 = (780 × 103)1−1.5 × T21.5
T21.5 = (150 × 103)1−1.5 × (300)1.5 × 3001.5
T21.5 = 11849.050
T2 = (11849.050)1/1.5
T2 = 519.74 = 520 K

(d) dQ = dW + dU
Or dW = −dU  [ Since dQ = 0 in an adiabatic process]
dW = −nCvdT
dW = −0.009 × 16.6 × (520 − 300)
dW = −0.009 × 16.6 × 220
dW = −32.87 J $\approx$ −33 J

(e)
dU = nCvdT
dU = 0.009 × 16.6 × 220 $\approx$ 33 J

#### Page No 79:

There are three gases A, B and C.
It is given that initially,
VA = VB = VC    and     TA = TB = TC .

For A, the process is isothermal and for an isothermal process, PV = constant.
PAVA = P'A2VA
$⇒P{\mathit{\text{'}}}_{\mathrm{A}}\mathit{=}{P}_{\mathrm{A}}\mathit{×}\frac{1}{2}$

For B, the process is adiabatic. So,
PBVBγ = PB'(2VB)γ
$⇒{P}_{\mathrm{B}}\text{'}\mathit{=}\frac{{P}_{\mathrm{B}}}{{2}^{1.5}}$

For C, the process is isobaric, which implies that the pressure will remain constant.
So, using the ideal gas equation
P$=\frac{\mathit{n}\mathit{R}\mathit{T}}{\mathit{V}}$, we get
$\frac{{V}_{\mathrm{C}}}{{T}_{\mathrm{C}}}\mathit{=}\frac{V{\mathit{\text{'}}}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathrm{C}}}\mathit{⇒}\frac{{V}_{\mathrm{C}}}{{T}_{\mathrm{C}}}\mathit{=}\frac{2{V}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathit{C}}}$
TC' = 2TC

As the final pressures are equal,

#### Page No 79:

Let,
Initial pressure of the gas = P1
Initial volume of the gas = V1
Final pressure of the gas= P2
Final volume of the gas = V2
Given, V2 = 2 V1, for each case.
In an isothermal expansion process,
work done
$W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}$
It is given that same work is done in both cases.
So,

${P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma }={P}_{1}{\left(\frac{1}{2}\right)}^{\gamma }$
From eq (1),

and nRT1 = P1V1

Or (γ − 1) ln 2 = 1 − 21−γ

#### Page No 79:

Given:
γ = 1.5
T = 300 K
Initial volume of the gas, V1 = 1 L
Final volume, V2 = $\frac{1}{2}$ L

(a) The process is adiabatic because volume is suddenly changed; so, no heat exchange is allowed.
P1V1γ = P2V2γ

(b) P1 = 100 kPa = 105 Pa
and P2 = $2\sqrt{2}$ × 105 Pa
Work done by an adiabatic process,

(c) Internal energy,
dQ = 0, as it is an adiabatic process.
⇒ dU = − dW = − (− 82 J) = 82 J

(d)
T1V1γ−1 = T2V2γ−1
T2 = T1${\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}$
=
300 $×$ (2)0.5
= 300 × $\sqrt{2}$ × = 300 × 1.4142
T2 = 424 K

(e) The pressure is kept constant.
The process is isobaric; so, work done = P$∆V$=nRdT.
Here, $n=\frac{PV}{RT}=\frac{{10}^{5}×{10}^{-3}}{R×300}=\frac{1}{3R}$
So, work done =
As pressure is constant,

(f)Work done in an isothermal process,

= 100 × ln 2 = 100 × 1.039
= 103 J

(g) Net work done (using first law of thermodynamics)
= − 82 − 41.4 + 103
= − 20.4 J

#### Page No 79:

Given:
γ = 1.5
For an adiabatic process, TVγ−1 = constant .
So, T1 V1γ−1 = T2 V2γ−1
As it is an adiabatic process and all the other conditions are same, the above equation can be applied.
In the new position, the slid is dividing the tube in the ratio 3:1.
So, if the total volume is V, then one side will occupy a volume of and the other side will occupy $\frac{V}{4}$.
So, ${T}_{1}×{\left(\frac{3v}{4}\right)}^{\gamma -1}={T}_{2}×{\left(\frac{v}{4}\right)}^{\gamma -1}$

(i) (ii) $⇒{T}_{1}×{\left(\frac{3v}{4}\right)}^{1.5-1}={T}_{2}×{\left(\frac{v}{4}\right)}^{1.5-1}\phantom{\rule{0ex}{0ex}}⇒\frac{{T}_{1}}{{T}_{2}}=\frac{\sqrt{3}}{1}$

#### Page No 79:

Given:
Volume of gas in each vessel, V = 200 cm3
Specific heat at constant volume of the gas, Cv = 12.5 J/mol-K
Initial temperature of the gas, T = 300 K
Initial pressure of the gas, P = 75 cm of Hg

(a) Using the ideal gas equation, number of moles of gases in each vessel,

(b) Heat is supplied to the gas, but dV is zero as the container has rigid walls.
So, dW = $P\mathit{∆}V$ = 0

From first law of thermodynamics,
dQ = dU
⇒ 5 = nCvdT
⇒ 5 = 0.008 × 12.5 × dT
dT = 50 for A
$\frac{\mathit{P}}{\mathit{T}}=\frac{{P}_{\mathrm{A}}}{{T}_{\mathrm{A}}}$ because volume is kept constant.
$Q=n{C}_{v}dT\phantom{\rule{0ex}{0ex}}T=\frac{Q}{n{C}_{v}}$

PB = 25 cm of Hg

The distance moved by the mercury,
PB − PA = 25 − 12.5 = 12.5 cm

#### Page No 79:

Given:
Mass of He, mHe = 0.1 g
γ1 = 1.67
Molecular weight of He, MHe = 4 g/mol
MH2 = ?
MH2 = 2 g/mol
γ2 = 1.4

Since it is an adiabatic environment and the system is not dong any external work, the amount of heat given will be used up entirely to raise its internal energy.
For He, dQ = dU = nCvdT   ...(i)
$=\frac{m}{4}×\frac{\mathrm{R}}{\gamma -1}×d\mathrm{T}\phantom{\rule{0ex}{0ex}}=\frac{0.1}{4}×\frac{\mathrm{R}}{\left(1.67-1\right)}×d\mathrm{T}$
For H2, dQ =dU = nCvdT   ...(ii)
$=\frac{m}{2}×\frac{\mathrm{R}}{\gamma -1}×d\mathrm{T}\phantom{\rule{0ex}{0ex}}=\frac{m}{2}×\frac{\mathrm{R}}{1.4-1}×d\mathrm{T},$

where m is the required mass of H2.

Since equal amount of heat is given to both gases; so dQ is same in both eq (i) and (ii), we get

#### Page No 79:

Initial pressure of the gas in both the vessels = P0
Initial temperature of the gas in both the vessels = T0
Initial volume = V0
$\frac{\mathrm{C}p}{\mathrm{C}v}=\gamma$

(a) The temperature inside the diathermic vessel remains constant. Thus,
P1V1 = P2V2
P0V0 = P2 × 2V0
$⇒{P}_{2}=\frac{{P}_{0}}{2}$
Temperature = T0
The temperature inside the adiabatic vessel does not remain constant.
T1V1γ−1 = T2V2γ−1
T0V0γ−1 = T × (2V0)γ−1
T2 = T0 × 21−γ
P1V1γ = P2V2γ
P0V0γ = P2 × (2V0)γ
$⇒{P}_{2}=\left(\frac{{P}_{0}}{{2}^{\gamma }}\right)$

(b) When the valves are open, the temperature remains T0 throughout, i.e. T1T2 = T0 .

Also, pressure will be same throughout. Thus, P1 = P2. As, temperature has not changed on side 1, so pressure on this side will also not change (volume is also fixed due to fixed piston) and will be equal to ${P}_{0}$ (pressure is an intrinsic variable).
On side 2, pressure will change to accommodate the changes in temperature on this side.
So, P0 = P1 + P2
2P1 = 2P2
So, P1 = P2 = $\frac{{\mathrm{P}}_{0}}{2}$

#### Page No 79:

For an adiabatic process, PVγ = Constant
So,  P1V1γ = P2V2γ   ...(i)
According to the problem,
V1 + V2 = V0  ...(ii)
Using the relation in eq (ii) in eq (i), we get
P1V1γ = P2(V0V1)γ

Using equation (ii), we get
${V}_{2}\mathit{=}\frac{{{P}_{1}}^{\frac{1}{\gamma }}{V}_{\mathit{0}}}{{{P}_{1}}^{\frac{1}{\gamma }}+{{P}_{2}}^{\frac{1}{\gamma }}}$

(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.
Hence, heat given to the gas in the left part is 0.

(c) There will be a common pressure 'P' when equilibrium is reached. The slid will move until the pressure on the two sides becomes equal.
P1V1γ + P2V2γ = PV0γ
For equilibrium, V1 = V2 = $\frac{{\mathrm{V}}_{0}}{2}$
Hence,

#### Page No 79:

Given:
Area of the tube, A = 1 cm2 = 1 × 10−4 m2
Mass of the gas, M = 0.03 g = 0.03 × 10−3 kg
Initial pressure, P = 1 atm = 105 Pascal
Initial length of the mercury column, L = 40 cm = 0.4 m
Final length of the mercury column, L1 = 80 cm = 0.8 m
Final pressure, P' = 0.355 atm

P(V)γ = P'(V')γ
⇒ 1 × (A × 0.4)γ = 0.355 × (A × 0.8)γ
⇒ 1 × 1 = 0.355 × 2γ

γ = 1.4941

Speed of sound in the gas,

#### Page No 80:

Given:
Velocity of sound in hydrogen, V = 1280 m/s
Temperature, T = 0°C = 273 K

Density of H2 = 0.089 kg/m3
R = 8.3 J/mol-K
At STP,
P = 105 Pa
We know:

#### Page No 80:

Given:
Specific heat capacity at constant pressure, Cp = 5.0 cal/mol-K
Cp = 5.0 × 4.2 J/mol-K
Cp = 21 J/mol-K
Volume of helium, V = 22400 ${\mathrm{cm}}^{3}$ = 0.0224 ${\mathrm{m}}^{3}$
At STP, P = 1 atm =
The speed of sound in gas,
$v\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{p}}{\mathit{\rho }}}\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{R}\mathit{T}}{\mathit{M}}}\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{P}\mathit{V}}{\mathit{M}}}\phantom{\rule{0ex}{0ex}}{C}_{\mathit{p}}\mathit{=}\frac{\mathit{R}\mathit{\gamma }}{\mathit{\gamma }\mathit{-}\mathit{1}}$
Or 21(γ − 1) = 8.3γ
⇒ 21γ − 8.3γ = 21
⇒ 12.7γ = 21

#### Page No 80:

Given:
Density of the ideal gas, ρ = 1.7 × 10−3 g/cm3
= 1.7 k/gm3
Pressure of the gas, P = 1.5 × 105 Pa
R = 8.3 J/mol-K

Resonance frequency of the gas = 3.0 kHz
Node separation in the Kundt's tube
= $\frac{l}{2}$ = 6 cm
So, l = 2$×$6 = 12 cm = 12 × 10−2 m
So, V = fl = 3 × 103 × 12 × 10−2
= 360 m/s
Speed of sound, V = $\sqrt{\frac{\gamma p}{\rho }}$
Or ${V}^{2}=\frac{\gamma p}{\mathrm{\rho }}$

#### Page No 80:

Frequency of standing waves, f = 5 × 103 Hz
Temperature of oxygen, T = 300 K
From Kundt's tube theory, we know that
$\frac{l}{2}$ = node separation = 3.3 cm
∴ l = 6.6 × 10−2 m
Also,
v = fl = 5 × 103 × 6.6 × 10−2
= (66 × 5) = 330 m/s

= 20.7 J mol-K

Specific heat at constant pressure, Cp = Cv + R
Cp = 20.7 + 8.3
Cp = 29.0 J/mol-K

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