Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in the given figure. By measuring the gradient, it can be shown that the gas may be
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a
monatomic and undergoing an adiabatic change
b
monatomic and undergoing an isothermal change
c
diatomic and undergoing an adiabatic change
d
triatomic and undergoing an isothermal change
answer is C.
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Detailed Solution
log P=m log V+C1, where C1 is positive, m is slopem=2.38−2.101.1−1.3=−1.4logP=−1.4logV+C1log PV1.4=C1PV1.4=kThus, it represents an ideal diatomic gas undergoing adiabatic change.