One mole of a monoatomic gas at pressure $2 atm,$ $279 K$ taken to final pressure $4 atm$ by a reversible path described by $\frac{P}{V} = constant$ . Calculate the magnitude of $\frac{Δ E}{w}$ for the process.

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Solution

We know that,
$d E = d q + d w P_{1} = K^{'} V_{1} and P_{2} = K^{'} V_{2} Also V_{1} = \frac{P_{1}}{K^{'}}, V_{2} = \frac{P_{2}}{K^{'}} ∴ T_{2} = T_{1} (\frac{P_{2} V_{2}}{P_{1} V_{1}}) \Rightarrow T_{2} = T_{1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{P_{2} P_{2}}{K^{'}}}{\frac{P_{2} P_{1}}{K^{'}}} ⎞ ⎟ ⎟ ⎟ ⎠ \Rightarrow T_{2} = T_{1} \frac{P_{2}^{2}}{P_{1}^{2}} \Rightarrow T_{2} = T_{1} \frac{4^{2}}{2^{2}} = 4 T_{1} Δ E = C_{V} (Δ T) = C_{V} (4 T_{1} - T_{1}) = 3 C_{V} T_{1} = 3 \frac{3}{2} R T_{1} = 9 \frac{R T_{1}}{2} (For monoatomic gas, C_{v} = \frac{3}{2} R) w = - \int_{V_{1}}^{V_{2}} P d V = - \int_{V_{1}}^{V_{2}} K^{'} V d V = - {∣ ∣ ∣ \frac{K^{'} V^{2}}{2} ∣ ∣ ∣}_{V_{1}}^{V_{2}} (K^{'} V^{2} = K^{'} V V = P V = R T) \Rightarrow w = - {∣ ∣ ∣ \frac{R T}{2} ∣ ∣ ∣}_{T_{1}}^{T_{2}} = - \frac{R}{2} (T_{2} - T_{1}) = - \frac{R}{2} 3 T_{1} \frac{Δ E}{w} = \frac{\frac{9 R T_{1}}{2}}{\frac{- 3 R T_{1}}{2}} = - 3$

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