First slide

Thermodynamic processes

Question

At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume $γ = 1.41$ . The work done on gas will be

Moderate

A

1280 J
B

1610 J
C

1815 J
D

2025 J

Solution

$\begin{array}{l} T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = 273 (2)^{0 .41} = 273 \times 1 .328 = 363 K \\ W = \frac{R (T_{1} - T_{2})}{γ - 1} = \frac{8 .31 (273 - 363)}{1 .41 - 1} = - 1824 \\ \Rightarrow | W | \approx 1815 J \end{array}$

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