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Internal Energy

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Question

$1$ $m o l e$ of ${C O}_{2}$ gas at $300 K$ is expanded under reversible adiabatic condition such that its volume becomes $27$ times. (a) What is the final temperature ? (b) What is work done ? Given $γ = 1.33$ and $C_{v} = 25.08 J {m o l}^{- 1} K^{- 1}$ for ${C O}_{2}$

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Solution

a) Using Poison's equation
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} o r \frac{T_{1}}{T_{2}} = (\frac{V_{2}}{V_{1}})^{γ - 1}$
Given $γ = 1.33$ So, $γ - 1 = 1.33 - 1 = 0.33$
$V_{2} = 27 V_{1} T_{1} = 300 K$
So, $\frac{300 K}{T_{2}} = (\frac{27 V_{1}}{V_{1}})^{0.33} T_{2} = 101.1 K$
Thus final temperature is $101.1 K$
b) For adiabatic condition
$d U = d Q + d W, d Q = 0$ for adiabatic
So, $W = d U W = C_{V} d T$
Given $C_{V} = 25.08 J m o l^{- 1} K^{- 1} T_{1} = 300 K T_{2} = 101.1 K W = 25.08 J m o l^{- 1} K^{- 1} (101.1 K - 300 K) W = - 4988.2 J m o l e^{- 1} o r - 4.98 K J m o l^{- 1}$
Thus work done is $- 4.98 K J m o l^{- 1}$

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