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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
1111062_07abe110fb8f4163bd34506d0935111f.png


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