An ideal gas at 27^{circ} C is compressed adi | vedclass

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(D) For an adiabatic process,the relation between temperature and volume is $TV^{\gamma-1} = 	ext{constant}$.Thus,$\frac{T_2}{T_1} = \left(\frac{V_1}

Vedclass · Accepted Answer

$375$

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(D) For an adiabatic process,the relation between temperature and volume is $TV^{\gamma-1} = 	ext{constant}$.Thus,$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}ight)^{\gamma-1}$.Given $T_1 = 27^{\circ} C = 300 \ K$ and $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.Given $\gamma = 5/3$,so $\gamma - 1 = 5/3 - 1 = 2/3$.Substituting the values: $\frac{T_2}{300} = \left(\frac{27}{8}ight)^{2/3} = \left(\left(\frac{3}{2}ight)^3ight)^{2/3} = \left(\frac{3}{2}ight)^2 = \frac{9}{4}$.$T_2 = \frac{9}{4} 	imes 300 = 9 	imes 75 = 675 \ K$.The rise in temperature is $\Delta T = T_2 - T_1 = 675 \ K - 300 \ K = 375 \ K$.

An ideal gas at $27^{\circ} C$ is compressed adiabatically to $8/27$ of its original volume. If the ratio of specific heats is $\gamma = 5/3$,then the rise in temperature of the gas is: (in $K$)

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