Initial pressure and volume of a monoatomic i | vedclass

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(C) For an adiabatic process,$PV^\gamma = 	ext{constant}$.For a monoatomic gas,the adiabatic index $\gamma = 5/3$.Given initial state: $P_1 = P, V_1

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$-\frac{4}{3}PV$

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(C) For an adiabatic process,$PV^\gamma = 	ext{constant}$.For a monoatomic gas,the adiabatic index $\gamma = 5/3$.Given initial state: $P_1 = P, V_1 = V$.Final volume: $V_2 = 27V$.Using the adiabatic relation $P_1V_1^\gamma = P_2V_2^\gamma$,we find the final pressure $P_2$:$P_2 = P(V_1/V_2)^\gamma = P(V/27V)^{5/3} = P(1/27)^{5/3} = P(1/3^3)^{5/3} = P(1/3^5) = P/243$.The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = \frac{f}{2} (P_2V_2 - P_1V_1)$,where $f$ is the degrees of freedom.For a monoatomic gas,$f = 3$.Substituting the values:$\Delta U = \frac{3}{2} ((\frac{P}{243}) \cdot 27V - PV) = \frac{3}{2} (\frac{P}{9} - PV) = \frac{3}{2} (\frac{P - 9P}{9}) = \frac{3}{2} (\frac{-8P}{9}) = -\frac{4}{3}PV$.

Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{final} = 27V$ is . . . . . . $J$.

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