QJEEAdv11Ph1255: JEE Advanced 2011 Physics - Class 11 JEEAdv11Ph1255 Solution: One mole of a monatomic gas is taken through a cycle ABCDA as…

Process A $\to$ B,

It is a isobaric process

P = constant, V $\propto$ T

$\because$ V_B < V_A $\Rightarrow$ T_B < T_A

$\Delta$ U = nC_V $\Delta$ T = $-$ ve

Hence internal energy decreases.

$\Delta$ Q = nC_P $\Delta$ T = $-$ ve

Hence heat is lost.

$\Delta$ W = nR $\Delta$ T = $-$ ve

Hence, work is done on the gas.

Process, B $\to$ C,

It is a isochoric process.

V = constant, P $\propto$ T

$\because$ P_C < P_B $\Rightarrow$ T_C < T_B

$\Delta$ U = nC_V $\Delta$ T = $-$ ve

Hence, internal energy decreases.

$\Delta$ W = 0, $\Delta$ Q = $\Delta$ U = $-$ ve

Hence, heat is lost.

Process C $\to$ D,

It is a isobaric process.

P = constant, V $\propto$ T

$\because$ V_D > V_C $\Rightarrow$ T_D > T_C

$\Delta$ U = nC_V $\Delta$ T = +ve

Hence internal energy increases.

$\Delta$ Q = nC_P $\Delta$ T = +ve

Hence, heat is gained.

$\Delta$ W = nR $\Delta$ T = +ve

Hence, work is done by the gas.

Process, D $\to$ A

According to ideal gas equation

${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$

Hence, it is a isothermal process.

$\therefore$ $\Delta$ U = 0

$\because$ V_A < V_D

$\Delta$ W = $-$ ve, hence work done is done on the gas.

$\Delta$ Q = $\Delta$ W = $-$ ve, hence heat is lost.