An ideal monoatomic gas of volume V is adiaba | vedclass

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(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.Given,initial volu

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$144.2$

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(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.Given,initial volume $V_1 = V$,final volume $V_2 = 3V$,and initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 	ext{ K}$.The adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$.Using the relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,we get:$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}ight)^{\gamma-1}$$\frac{T_2}{300} = \left(\frac{V}{3V}ight)^{\frac{5}{3}-1} = \left(\frac{1}{3}ight)^{\frac{2}{3}}$$T_2 = 300 	imes (3)^{-\frac{2}{3}} = 300 	imes \frac{1}{3^{0.666}} \approx 300 	imes 0.4807 \approx 144.2 	ext{ K}$.

An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3V$ at $27^{\circ}C$. The final temperature in Kelvins is (use $\frac{C_P}{C_V} = \frac{5}{3}$)

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