Question

Four moles of an ideal gas (monatomic) at pressure $P_{0}$ and temperature $T_{0}$ is isothermally expanded to twice its initial volume. The gas is then compressed at constant pressure to its original volume. Finally, the gas is compressed at constant volume to its original pressure $A \to B$ .

Sketch the $W_{A B} = n R T \ln (\frac{V_{B}}{V_{A}}) = 4 R T_{0} \ln 2$ and $P - T$ diagrams for the process.

Sketch the $P - V$ and $P - T$ diagrams for the process.

Accepted Answer

Let the initial and the two successive intermediate states be symbolised by $A, B$ and $C$ , respectively.

For state $A$ : $P_{A} = P_{0}; T_{A} = T_{0}$ and $V_{A} = \frac{4 R T_{0}}{P_{0}}$

$(∵ P V = n R T)$

For state $B$ : $V_{B} = \frac{8 R T_{0}}{P_{0}}; T_{B} = T_{0}$

(the process $A \to B$ being isothermal)

and $P_{B} = \frac{P_{0}}{2}$

For state $C$ : $V_{C} = \frac{4 R T_{0}}{P_{0}}; P_{C} = \frac{P_{0}}{2}$

(the process $B \to C$ being isobaric)

$∴ V \propto T$ and $T_{C} = \frac{T_{0}}{2}$

The $P - V$ and $P - T$ curves are shown in Figs. (a) and (b).

a

b

(b) For the process $A \to B$ (isothermal):

Work done, $W_{A B} = n R T \ln (\frac{V_{B}}{V_{A}}) = 4 R T_{0} \ln 2$

For the process $B \to C$ (isobaric):

$W_{B C} = P Δ T = n R Δ T = 4 R (\frac{T_{0}}{2} - T_{0}) = - 2 R T_{0}$

The work done in part $C \to A$ is zero as the volume remains constant. Hence, $W_{c A} = 0$ .

Thus net work done during the complete cycle,

$W = W_{A B} + W_{B C} + W_{C A}$

$= 4 R T_{0} \ln 2 - 2 R T_{0} + 0 = 2 R T_{0} (2 \ln 2 - 1)$

As the process is cyclic, the change in internal energy is zero. The heat given to the system is then equal to the work done by it.

i.e., heat given to the gas during the complete cycle,

$Q = 2 R T_{0} (2 \ln 2 - 1)$

B M Sharma Solutions for Chapter: Thermodynamics, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Thermodynamics, Exercise 1: ILLUSTRATIONS

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