Starting at temperature 300 ; K, one mole of | vedclass

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(A) Step $1$: Adiabatic compression from $V_{1}$ to $V_{2} = V_{1}/16$.For an adiabatic process,$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$.Given

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$1818$

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(A) Step $1$: Adiabatic compression from $V_{1}$ to $V_{2} = V_{1}/16$.For an adiabatic process,$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$.Given $T_{1} = 300 \; K$,$\gamma = 1.4 = 7/5$,so $\gamma-1 = 0.4 = 2/5$.$300 	imes V_{1}^{2/5} = T_{2} 	imes (V_{1}/16)^{2/5}$.$T_{2} = 300 	imes (16)^{2/5} = 300 	imes (2^{4})^{2/5} = 300 	imes 2^{8/5}$.$T_{2} = 300 	imes 3.0314 = 909.42 \; K$.Step $2$: Isobaric expansion from $V_{2}$ to $2V_{2}$.For an isobaric process,$V/T = 	ext{constant}$,so $V_{2}/T_{2} = (2V_{2})/T_{f}$.$T_{f} = 2 	imes T_{2} = 2 	imes 909.42 = 1818.84 \; K$.Rounding to the nearest integer,the final temperature is $1819 \; K$. Given the options,$1818$ is the closest match.

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