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Thermodynamic Processes

A cyclic proc...

Question

A cyclic process for $1 m o l e$ of an ideal gas is shown in the $V - T$ diagram. The work done in $A B, B C$ and $C A$ respectively is

0, T R_{2} ln ∣ ∣ ∣ \frac{V_{2}}{V_{1}} ∣ ∣ ∣, R (T_{1} - T_{2})

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R (T_{1} - T_{2}), 0, R T_{1} ln ∣ ∣ ∣ \frac{V_{1}}{V_{2}} ∣ ∣ ∣

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0, R T_{1} ln ∣ ∣ ∣ \frac{V_{2}}{V_{1}} ∣ ∣ ∣, \frac{R T_{1}}{V_{1}} (T_{1} - T_{2})

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0, R T_{2} ln ∣ ∣ ∣ \frac{V_{2}}{V_{1}} ∣ ∣ ∣, R (T_{1} - T_{2})

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Solution

The correct option is D $0, R T_{2} ln ∣ ∣ ∣ \frac{V_{2}}{V_{1}} ∣ ∣ ∣, R (T_{1} - T_{2})$
During $A B$ , process is isochoric
$∴ Δ V = 0$
$∴ W = 0$
During $B C$ , process is isothermal
$∴ Δ T = 0$
$∴ W = R T_{2} l n \frac{V_{2}}{V_{1}}$
During $C A$ , process is isobaric. So, pressure is constant
$∴ W = P (V_{1} - V_{2})$
But $P V_{1} = R T_{1}$
$∴ P = \frac{R T_{1}}{V_{1}} = \frac{R T_{2}}{V_{2}}$

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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

The correct option is D 0,RT2ln∣∣∣V2V1∣∣∣,R(T1−T2)During AB, process is isochoric∴ΔV=0∴W=0During BC, process is isothermal∴ΔT=0∴W=RT2lnV2V1During CA, process is isobaric. So, pressure is constant∴W=P(V1−V2)But PV1=RT1∴P=RT1V1=RT2V2

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