10 exam-style questions on ideal gases, kinetic theory, specific heat capacity and internal energy. Tap Show answer for full working.
A fixed mass of ideal gas occupies \(0.025\,\mathrm{m}^{3}\) at \(300\,\mathrm{K}\) and pressure \(1.2\times10^{5}\,\mathrm{Pa}\). Calculate the number of moles. (\(R = 8.31\,\mathrm{J\,mol}^{-1}\mathrm{K}^{-1}\).)
\(pV = nRT \Rightarrow n = pV/RT = (1.2\times10^{5}\times0.025)/(8.31\times300) = 1.20\,\mathrm{mol}\).
Calculate the rms speed of oxygen molecules (\(M = 32\,\mathrm{g\,mol}^{-1}\)) at \(27^\circ\mathrm{C}\). (\(R = 8.31\,\mathrm{J\,mol}^{-1}\mathrm{K}^{-1}\).)
\(c_{\mathrm{rms}} = \sqrt{3RT/M} = \sqrt{3\times8.31\times300/0.032} = 483\,\mathrm{m\,s}^{-1}\).
How much energy is required to raise the temperature of \(2.0\,\mathrm{kg}\) of water by \(15\,\mathrm{K}\)? (Specific heat capacity of water \(= 4200\,\mathrm{J\,kg}^{-1}\mathrm{K}^{-1}\).)
\(Q = mc\Delta\theta = 2.0\times4200\times15 = 1.26\times10^{5}\,\mathrm{J}\).
An ideal gas expands isothermally from \(0.010\,\mathrm{m}^{3}\) to \(0.025\,\mathrm{m}^{3}\) at \(350\,\mathrm{K}\). The initial pressure is \(2.0\times10^{5}\,\mathrm{Pa}\). Calculate the final pressure and the work done by the gas. (\(R = 8.31\,\mathrm{J\,mol}^{-1}\mathrm{K}^{-1}\).)
\(p_{1}V_{1} = p_{2}V_{2} \Rightarrow p_{2} = 2.0\times10^{5}\times0.010/0.025 = 8.0\times10^{4}\,\mathrm{Pa}\).
\(n = p_{1}V_{1}/RT = (2\times10^{5}\times0.01)/(8.31\times350) = 0.688\,\mathrm{mol}\).
\(W = nRT\ln(V_{2}/V_{1}) = 0.688\times8.31\times350\times\ln(2.5) = 1830\,\mathrm{J}\).
State the first law of thermodynamics. A gas absorbs \(500\,\mathrm{J}\) of heat and expands, doing \(200\,\mathrm{J}\) of work. Calculate the change in internal energy.
\(\Delta U = Q - W\) (or \(Q + W\) depending on sign convention; here \(W\) is work done by the system).
\(\Delta U = 500 - 200 = 300\,\mathrm{J}\).
A sample of ideal gas has pressure \(p\), volume \(V\) and temperature \(T\). The volume is doubled while the temperature is held constant. State what happens to the pressure and to the mean kinetic energy of the molecules. Explain your answers.
Pressure halves (\(pV = \mathrm{const}\) at constant \(T\)).
Mean kinetic energy unchanged (depends only on temperature: \(\frac12 m\langle c^{2}\rangle = \frac32 kT\)).
Calculate the internal energy of \(3.0\,\mathrm{mol}\) of monatomic ideal gas at \(400\,\mathrm{K}\). (\(R = 8.31\,\mathrm{J\,mol}^{-1}\mathrm{K}^{-1}\).) If the gas is heated at constant volume to \(500\,\mathrm{K}\), how much heat is required?
For monatomic gas \(U = \frac32 nRT = 1.5\times3\times8.31\times400 = 14960\,\mathrm{J}\).
At constant volume \(Q = \Delta U = \frac32 nR\Delta T = 1.5\times3\times8.31\times100 = 3740\,\mathrm{J}\).
Derive the relation \(\frac12 m\langle c^{2}\rangle = \frac32 kT\) from the kinetic theory expression \(p = \frac13\rho\langle c^{2}\rangle\) and the ideal gas law. Hence calculate the mean kinetic energy of a molecule at \(300\,\mathrm{K}\). (\(k = 1.38\times10^{-23}\,\mathrm{J\,K}^{-1}\).)
\(pV = \frac13 Nm\langle c^{2}\rangle\) and \(pV = NkT \Rightarrow \frac13 m\langle c^{2}\rangle = kT \Rightarrow \frac12 m\langle c^{2}\rangle = \frac32 kT\).
\(\langle KE\rangle = \frac32\times1.38\times10^{-23}\times300 = 6.21\times10^{-21}\,\mathrm{J}\).
A \(0.50\,\mathrm{kg}\) block of copper at \(100^\circ\mathrm{C}\) is placed in \(1.2\,\mathrm{kg}\) of water at \(20^\circ\mathrm{C}\) in an insulated container. Calculate the final equilibrium temperature. (Specific heat capacities: copper \(390\,\mathrm{J\,kg}^{-1}\mathrm{K}^{-1}\), water \(4200\,\mathrm{J\,kg}^{-1}\mathrm{K}^{-1}\).)
Heat lost by copper = heat gained by water.
\(0.50\times390\times(100-\theta) = 1.2\times4200\times(\theta-20)\).
\(195(100-\theta) = 5040(\theta-20)\).
\(19500 - 195\theta = 5040\theta - 100800\).
\(120300 = 5235\theta \Rightarrow \theta = 23.0^\circ\mathrm{C}\).
An ideal gas is taken through a cycle consisting of an isochoric pressure increase, an isothermal expansion, and an isobaric compression back to the original state. Sketch the \(p\)–\(V\) diagram and indicate on which parts of the cycle the internal energy increases, decreases or stays constant.
Vertical line up (isochoric heating: \(U\) increases), horizontal curve to the right (isothermal expansion: \(U\) constant), vertical-ish or horizontal left (isobaric compression: volume decreases, \(T\) falls so \(U\) decreases).
Internal energy of an ideal gas depends only on temperature: increases when \(T\) rises, constant in isothermal process, decreases when \(T\) falls.