ThermodynamicsMedium

Question

An ideal monoatomic gas undergoes a reversible process, where $\frac{P}{V}$ = constant, from (2 bar, 273 K) to 4 bar. The value of $\frac{\Delta U}{w}$ for this process is

Options

A.+3.0
B.−3.0
C.−1.5
D.+1.5

Solution

$P.V^{- 1} = K \Rightarrow x = - 1$

$\text{Now, }\frac{\Delta U}{w} = \frac{n.C_{v,m}.\Delta T}{\left( \frac{n.R.\Delta T}{x - 1} \right)} = - 2.\frac{C_{v,m}}{R} = - 2 \times \frac{\frac{3}{2}R}{R} = - 3$

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