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Question

(A) Given: Initial temperature $T_i = 300 \ K$, Initial pressure $P_i = P$, Final pressure $P_f = 4P$, and the process relation $PV^{4/3} = 	ext{cons

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$300\sqrt{2}$

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(A) Given: Initial temperature $T_i = 300 \ K$, Initial pressure $P_i = P$, Final pressure $P_f = 4P$, and the process relation $PV^{4/3} = 	ext{constant}$.For an ideal gas, $V = \frac{nRT}{P}$.Substituting $V$ into the given relation: $P \left( \frac{nRT}{P} ight)^{4/3} = 	ext{constant}$.This simplifies to $P \cdot P^{-4/3} \cdot T^{4/3} = 	ext{constant}$, which gives $T^{4/3} \cdot P^{-1/3} = 	ext{constant}$.Thus, $\frac{T_f^{4/3}}{P_f^{1/3}} = \frac{T_i^{4/3}}{P_i^{1/3}}$.Rearranging for $T_f$: $T_f = T_i \left( \frac{P_f}{P_i} ight)^{(1/3) / (4/3)} = T_i \left( \frac{P_f}{P_i} ight)^{1/4}$.Substituting the values: $T_f = 300 	imes \left( \frac{4P}{P} ight)^{1/4} = 300 	imes (4)^{1/4} = 300 	imes (2^2)^{1/4} = 300 	imes 2^{1/2} = 300\sqrt{2} \ K$.

An ideal gas undergoes an adiabatic process obeying the relation $PV^{4/3} = \text{constant}$. If its initial temperature is $300 \ K$ and its pressure is increased up to four times its initial value, then the final temperature is (in Kelvin):

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