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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…

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$1.2\ \text{mol}$ of an ideal gas at $310\ \text{K}$ is squeezed slowly at a steady temperature from $24$ to $6.0$ litres. Find the work done by the gas, its change of internal energy, the heat, and its first and last pressures. ( $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ )

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: $W = 0$ , $Q = \Delta U = nC_v\,\Delta T$ .

Work done by a gas is the area under its path on a pressure-volume graph. At a steady pressure $W = p\,\Delta V$ ; along a straight line, the average pressure times $\Delta V$ ; at a steady temperature $W = nRT\ln\dfrac{V_2}{V_1}$ . Expanding gives positive work, squeezing negative. $1\ \text{L} = 10^{-3}\ \text{m}^3$ .

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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