In an H_2 gas process, PV^2 = text{constant}. | vedclass

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(C) For a polytropic process $PV^x = 	ext{constant}$, the work done is given by $W = \frac{nR \Delta T}{1-x}$.Here, $x = 2$, so $W = \frac{nR \Delta

Vedclass · Accepted Answer

$-0.4$

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(C) For a polytropic process $PV^x = 	ext{constant}$, the work done is given by $W = \frac{nR \Delta T}{1-x}$.Here, $x = 2$, so $W = \frac{nR \Delta T}{1-2} = -nR \Delta T$.For a diatomic gas like $H_2$, the change in internal energy is $\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R ight) \Delta T = 2.5 nR \Delta T$.The ratio of work done to the change in internal energy is $\frac{W}{\Delta U} = \frac{-nR \Delta T}{2.5 nR \Delta T} = -\frac{1}{2.5} = -0.4$.

In an $H_2$ gas process, $PV^2 = \text{constant}$. The ratio of work done by the gas to the change in its internal energy is:

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