A monoatomic gas left(gamma=frac{5}{3}right) | vedclass

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(D) For an adiabatic process,the relationship between pressure $p$ and temperature $T$ is given by $T^\gamma p^{1-\gamma} = 	ext{constant}$.Thus,$T_1

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$2^{\frac{9}{2}} 	ext{ atm}$

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(D) For an adiabatic process,the relationship between pressure $p$ and temperature $T$ is given by $T^\gamma p^{1-\gamma} = 	ext{constant}$.Thus,$T_1^\gamma p_1^{1-\gamma} = T_2^\gamma p_2^{1-\gamma}$.Rearranging for $p_2$,we get $p_2 = p_1 \left(\frac{T_1}{T_2}ight)^{\frac{\gamma}{1-\gamma}}$.Given values: $p_1 = 4 	ext{ atm} = 2^2 	ext{ atm}$,$\gamma = 5/3$,$T_1 = 27^{\circ} 	ext{C} = 300 	ext{ K}$,and $T_2 = 327^{\circ} 	ext{C} = 600 	ext{ K}$.Substituting these values:$p_2 = 2^2 \left(\frac{300}{600}ight)^{\frac{5/3}{1-5/3}} = 2^2 \left(\frac{1}{2}ight)^{\frac{5/3}{-2/3}} = 2^2 \left(\frac{1}{2}ight)^{-5/2} = 2^2 	imes 2^{5/2}$.$p_2 = 2^{2 + 5/2} = 2^{9/2} 	ext{ atm}$.

$A$ monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ at a pressure of $4 \text{ atm}$ is compressed adiabatically so that its temperature rises from $27^{\circ} \text{C}$ to $327^{\circ} \text{C}$. The pressure of the gas in its final state is

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