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The slope of the graph between log P and log V at constant temperature for a given mass of a gas is:-

Kindly answer sir/mam.

a) +1

b) -1

c) 1/T

d) 1/n

$AccordingtoBoyle\text{'}slaw,\phantom{\rule{0ex}{0ex}}PV=cons\mathrm{tan}t\phantom{\rule{0ex}{0ex}}P=\frac{1}{V}cons\mathrm{tan}t\phantom{\rule{0ex}{0ex}}\mathrm{log}P=-\mathrm{log}V+cons\mathrm{tan}t\phantom{\rule{0ex}{0ex}}Whenweplot\mathrm{log}Pvs\mathrm{log}Vwegetanegativeslope=-1.$

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