A thermodynamic process for one mole of an id | vedclass

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(D) For a polytropic process, the equation is $PV^{x} = 	ext{constant}$.Given the process equation is $PV^{1/2} = 	ext{constant}$.Using the ideal ga

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$\frac{1}{\sqrt{2}}$

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(D) For a polytropic process, the equation is $PV^{x} = 	ext{constant}$.Given the process equation is $PV^{1/2} = 	ext{constant}$.Using the ideal gas equation $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting this into the process equation:$\left(\frac{nRT}{V}ight) V^{1/2} = 	ext{constant}$$T V^{-1} V^{1/2} = 	ext{constant}$$T V^{-1/2} = 	ext{constant}$Thus, $T_{1} V_{1}^{-1/2} = T_{2} V_{2}^{-1/2}$.Rearranging for the ratio of temperatures:$\frac{T_{2}}{T_{1}} = \left(\frac{V_{2}}{V_{1}}ight)^{-1/2} = \left(\frac{V_{1}}{V_{2}}ight)^{1/2}$.Given $V_{2} = 2V_{1}$, we have $\frac{V_{1}}{V_{2}} = \frac{1}{2}$.Therefore, $\frac{T_{2}}{T_{1}} = \left(\frac{1}{2}ight)^{1/2} = \frac{1}{\sqrt{2}}$.Hence, the correct option is $D$.

$A$ thermodynamic process for one mole of an ideal gas is shown below on a $P-V$ diagram. If $V_{2} = 2V_{1}$, then the ratio of temperature $T_{2} / T_{1}$ is ...... .

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