Calculate the number of moles of air in the cylinder.

The cork leaves the toy after the air is compressed to a pressure of 1.9×10⁵Pa and a volume of 4.0×10^–4 m³.

(i) Deduce that the compression of the gas is not isothermal.

(ii) Outline why the compression might be adiabatic.

(iii) The work needed to compress the air in (a) is 15J. Determine, with reference to the first law of thermodynamics, the change in the internal energy of the air in the cylinder.

(iv) Calculate the change in average kinetic energy of an air molecule as a result of the compression.

The piston is now pushed in slowly so that the compression is isothermal. Discuss the entropy changes that take place in the air of the toy and in its cylinder as the air is compressed.

\(n = \frac{{pV}}{{RT}} = \frac{{1.1 \times {{10}^5} \times 6.0 \times {{10}^{ - 4}}}}{{8.31 \times 290}}\);
0.027;

(i) calculate pV correctly for both states: 66 and 76; } (do not penalize 66 k/K and 76 k/K as k may be a constant)
isothermal change would mean that p₁V₁=p₂V₂;
so not isothermal

(ii) no heat/thermal energy transferred;
(because change/compression) occurs (too) quickly/fast; [2]

(iii) Q=0;
W=−15; (minus sign is required)
so ΔU=(+)15J; } (symbols must be defined)
(allow ECF from second marking point)

(iv) number of air molecules=0.0274×6.0×10²³(=1.64×10²²);
9.13×10^–22J;

entropy is a property that indicates degree of disorder in the system / OWTTE;
gas occupies a smaller volume; (do not allow “compressed”)
entropy of gas/air in toy decreases because more ordered;
energy reaches surroundings/cylinder / entropy cannot decrease;
so entropy of surroundings/cylinder increases; } (do not allow ECF from “entropy of gas increases”)