Single Choice

A cyclic process for $$1\ mole$$ of an ideal gas is shown in the $$V-T$$ diagram. The work done in $$AB, BC$$ and $$CA$$ respectively is

A$$0,\, TR_2\, \ln\, \left | \displaystyle \dfrac{V_2}{V_1} \right |,\, R(T_1\, -\, T_2)$$
B$$R(T_1\, -\, T_2),\, 0,\, RT_1\, \ln\, \left |\displaystyle \dfrac{V_1}{V_2} \right |$$
C$$0,\, RT_1\, \ln\, \left |\displaystyle \dfrac{V_2}{V_1} \right |, \dfrac{RT_1}{V_1}(T_1\, -\, T_2)$$
D$$0,\, RT_2\, \ln\, \left | \displaystyle \dfrac{V_2}{V_1} \right |,\, R(T_1\, -\, T_2)$$
Correct Answer

Solution

$$PV=nRT\\ P\propto T\\ V\propto T$$
Work Done , $$W=\int { PdV } $$

For Process A-B
Volume is constant and temperature is increasing.
Therefore, Work Done is Zero.

For Process B-C:
Volume is increasing with temperature being constant. So insentropic process.
Therefore, Work Done = $$R{ T }_{ 2 }\ \ln\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } $$

For Process C-A:
Volume is decreasing linearly with temperature. Therefore, P=cT kind of equation.
Therefore, Work Done = $$\int _{ V2 }^{ V1 }{ P } dV\quad =\quad R({ T }_{ 1 }-{ T }_{ 2 })$$


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