Question

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

A
0,RT1ln(V1V2),R(T1T2)
B
R,(T1T2)R,RT1ln(V1V2)
C
0,RT1ln(V2V1),RT1V1(V1V2)
D
0,RT2ln(V2V1),R(T2T1)

Solution

## The correct option is C $$0, RT_{1} ln \left (\dfrac {V_{2}}{V_{1}}\right ), \dfrac {RT_{1}}{V_{1}} (V_{1} - V_{2})$$$$\textbf{Explanation:-}$$During AB, process is isochoric$$\therefore \Delta V = 0$$        $$\therefore W = 0$$During BC, process is isothermal$$\therefore \Delta T = 0$$         $$\therefore W = RT_{2}ln\frac{V_{2}}{V_{1}}$$During CA, process is isobaric. So, pressure is constant,$$\therefore W = P(V_{1} - V_{2})$$But $$PV_{1} = RT_{1}$$$$\therefore P = \frac{RT_{1}}{V_{1}} = \frac{RT_{2}}{V_{2}}$$$$\therefore W = \frac{RT_{1}}{V_{1}}\left ( V_{1} - V_{2} \right ) = \frac{RT_{2}}{V_{2}}\left ( V_{1} - V_{2} \right )$$$$\textbf{Hence the correct option is C}$$Physics

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