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(A) For an ideal gas,the equation of state is $PV = nRT$,which can be written as $P = nRT(1/V)$.Along $AB$,the pressure $P$ is constant,and the value

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Along $AB$,temperature decreases while along $BC$ temperature increases.

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(A) For an ideal gas,the equation of state is $PV = nRT$,which can be written as $P = nRT(1/V)$.Along $AB$,the pressure $P$ is constant,and the value of $(1/V)$ increases. Since $P = nRT(1/V)$,if $P$ is constant and $(1/V)$ increases,the temperature $T$ must decrease.Along $BC$,the value of $(1/V)$ is constant,and the pressure $P$ increases. Since $P = nRT(1/V)$,if $(1/V)$ is constant and $P$ increases,the temperature $T$ must increase.Therefore,the correct statement is that along $AB$,temperature decreases while along $BC$ temperature increases.

An enclosed ideal gas is taken through a cycle as shown in the figure. Then

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