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

0.80 mol of a monatomic ideal gas goes from A to C in two steps.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

$0.80\ \text{mol}$ of a monatomic ideal gas goes from A to C in two steps. A to B: it is heated at a steady volume from $280\ \text{K}$ to $560\ \text{K}$ . B to C: it expands at a steady $560\ \text{K}$ to three times its volume. Find $W$ , $\Delta U$ and $Q$ for each step and for the whole change. ( $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ )

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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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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← 1.2 mol of an ideal gas at 310 K is squeezed slowly at a steady temperature from 24 to 6.0 litres.Air (γ = 1.40) at 1.0 × 10⁵ Pa and 290 K is squeezed quickly to 1/12 of its volume. →

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