# An lpg cylinder contains 15kg of butane at 27oC and 10 bar pressure. The cylinder leaked and it was found that after 4 days,the mass of the gas left in cylinder in was only 10 kg. How much pressure was dropped in each day (average pressure drop/day)?

According to ideal gas equation, PV = nRT

In the question, V, R and T remains constant. Only P and n are changing

P is directly proportional to n.

n = number of moles = mass/molar mass

Molar mass in both the cases is also constant, so. P is directly proportional to mass of gas.

$\frac{{P}_{1}}{{P}_{2}}=\frac{{m}_{1}}{{m}_{2}}$

P

_{1}= 10 bar, P

_{2}= ?

m

_{1}= 15 kg, m

_{2}= 10 kg

After putting the values,

$\frac{10}{{P}_{2}}=\frac{15}{10}$

P

_{2}= 6.67 bar

Pressure loss = 10-6.67 = 3.33 bar

Number of days = 4

Pressure drop per day = 3.33/4 = 0.83 bar/day

Regards

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