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P-T Relationship

One mole of m...

Question

One mole of monoatomic ideal gas at P = 2 bar and T = 273 K is compressed to 4 bar pressure following a reversible path obeying $\frac{P}{V} = c o n s t a n t$ . Assume $C_{v} = {12.5 J m o l}^{- 1} K^{- 1}$ . The value of $\frac{Δ u}{w}$ for this process is $x$ , then $- x$ is _____.

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Solution

P $=$ kV
When P is doubled, V is doubled (say) from V to 2V.
To obtain work done, kV is integrated with respect to V within the limits V, 2V.
Work done $W = 1.5 k V^{2}$ .....(1)
The ideal gas equation is
$P V = R T$
But $P = k V$ and $T = 273$ K
Hence, $k V^{2} = R (273)$ ....(2)
Substitute this in equation (1)
$W = 1.5 (R) (273)$
But $C_{v} = 1.5 R$
Hence, $W = C_{v} (273)$ ....(3)
In equation (2) when we substitute V with 2V and 273 with T, we get the final temperature $T = 4 (273)$
Change in internal energy $Δ U = C_{V} (T_{2} - T_{1}) U = C_{V} (273) (4 - 1) = 3 C_{V} (273)$ ...(4)
Divide equation (4) with equation (3).
$\frac{Δ U}{W} = \frac{3 C_{V} (273)}{C_{v} (273)} = 3$

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