A work of 166.28 J is done to adiabatically | vedclass

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(B) Work done on the system,$W = 166.28 \ J$. Increase in temperature,$\Delta T = 8^{\circ} C = 8 \ K$. For an adiabatic process,the work done on the

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diatomic

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(B) Work done on the system,$W = 166.28 \ J$. Increase in temperature,$\Delta T = 8^{\circ} C = 8 \ K$. For an adiabatic process,the work done on the gas is given by $W = \frac{nR\Delta T}{\gamma - 1}$. Given $n = 1 \ mol$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$. Substituting the values: $166.28 = \frac{1 	imes 8.314 	imes 8}{\gamma - 1}$. $\gamma - 1 = \frac{66.512}{166.28} = 0.4$. $\gamma = 1.4$. Since the adiabatic index $\gamma = 1.4$ corresponds to a diatomic gas,the gas is diatomic.

$A$ work of $166.28 \ J$ is done to adiabatically compress one mole of a gas. If the increase in the temperature of the gas is $8^{\circ} C$,the gas is $\left(R=8.314 \ J \ mol^{-1} \ K^{-1}\right)$

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