MediumPhysicsHigh School (11–12)Heat & Thermodynamics › Thermodynamic Processes and Specific Heats of GasesNumerical

0.40 mol of a diatomic ideal gas is at 290 K and 1.5 × 10⁵ Pa.

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$0.40\ \text{mol}$ of a diatomic ideal gas is at $290\ \text{K}$ and $1.5 \times 10^{5}\ \text{Pa}$ . It is heated at a steady pressure until its volume is $1.5$ times as large. Find its first volume, its final temperature, the work it does, its change of internal energy and the heat. ( $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ )

Need a reminder? The key ideaShow

Isothermal: $\Delta U = 0$ , $Q = W = nRT\ln\dfrac{V_2}{V_1}$ . Isobaric: $W = p\,\Delta V = nR\,\Delta T$ , $Q = nC_p\,\Delta T$ . Isochoric: $W = 0$ , $Q = \Delta U = nC_v\,\Delta T$ .

Molar heat capacities: $C_v = \tfrac{f}{2}R$ and $C_p = C_v + R$ (Mayer), with $\gamma = \dfrac{C_p}{C_v}$ . Monatomic: $12.5\ \text{J mol}^{-1}\text{K}^{-1}$ , $20.8\ \text{J mol}^{-1}\text{K}^{-1}$ , $\gamma = 1.67$ . Diatomic: $20.8\ \text{J mol}^{-1}\text{K}^{-1}$ , $29.1\ \text{J mol}^{-1}\text{K}^{-1}$ , $\gamma = 1.40$ . $Q = nC\,\Delta T$ , and $\Delta U = nC_v\,\Delta T$ in every process.

The worked solution

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