| Date | May 2011 | Marks available | 3 | Reference code | 11M.2.HL.TZ2.5 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate and State | Question number | 5 | Adapted from | N/A |
Question
This question is about changes of state of a gas.
A cylinder fitted with a piston contains 0.23 mol of helium gas.
The following data are available for the helium with the piston in the position shown.
Volume = 5.2×10–3m3
Pressure = 1.0 ×105 Pa
Temperature = 290K
Markscheme
(i) use of \(R = \frac{{pV}}{{nT}}\); (award mark if correct substitution seen)
\(\left( {\frac{{5.2 \times {{10}^{ - 3}} \times 1.0 \times {{10}^5}}}{{0.23 \times 290}}} \right) = 7.8{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\); (accept Pa m3 mol−1 K−1 )
(ii) the gas is ideal;
constant temperature required; (do not allow “isothermal”)
a slow compression allows time for (internal) energy to leave gas / OWTTE;
(for adiabatic change) Q=0;
W is positive / work is done by the gas;
∆U =−W so ∆U is negative;
(T is a measure of U therefore) T less than 290K;
Examiners report
(ii) Most recognized that the gas has to be ideal for the calculation in (i) to be carried through.
