Volume versus temperature graph of two moles of helium gas is as shown in Fig. The ratio of heat shown in Fig. The ratio of heat absorbed and work done by the gas in process 1-2 is :

3

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5 / 2

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5 / 3

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Solution

The correct option is A $5 / 2$
$d Q = n C_{v} d T$

$d W = n (C_{p} - C_{v}) d T$

$\frac{C_{p}}{C_{p} - C_{v}}$

$C_{v} / C_{p} = 3 / 5$

$C_{v} = R + 3 / 2 R$

Degree of freedom = 3

$d Q / d W = \frac{1}{1 - C_{p} / C_{v}}$

$C_{v} / C_{p} = 3 / 5$

$d Q / d W = \frac{1}{1 - 3 / 5}$

$d Q / d W = \frac{5}{2}$

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Volume versus temperature graph of two moles of helium gas is as shown in Fig. The ratio of heat shown in Fig. The ratio of heat absorbed and work done by the gas in process 1-2 is :

The correct option is A 5/2dQ=nCvdTdW=n(Cp−Cv)dTCpCp−CvCv/Cp=3/5Cv=R+3/2RDegree of freedom = 3dQ/dW=11−Cp/CvCv/Cp=3/5dQ/dW=11−3/5dQ/dW=52