Calculate the temperature of the gas at point C.

Compare, without any calculation, the work done and the thermal energy supplied along route ABC and route ADC.

use of \(\frac{{PV}}{T} = \) constant or use of \(T \propto PV\) or via intermediate calculation of \(n\) in \(PV = nRT\);

\(\frac{{1.95}}{{358}} = \frac{{8.55}}{{{T_{\text{c}}}}}\) or \(n = 6.55 \times {10^{ - 3}}{\text{ (mol)}}\); } (allow power of ten omission provided same omission on both sides)

1570 K or 1300 °C;

Award [3] for a bald correct answer ignoring small rounding errors.

Omitting conversion to Kelvin yields answer of 373 – award [2 max] as one error.

Award [1 max] if 273 subsequently added (to give 646).

same temperature change so same change in internal energy/\(\Delta U\);

work done along ABC is larger/ADB is smaller because area under ABC is greater than area under ADC/\(\Delta V\) same in both, \(P\) greater for ABC so \(P\Delta V\) also greater for ABC;

because \(\Delta Q = \Delta U + W\) thermal energy transferred is greater for route ABC/smaller for route ADB;

Must see reference to first law for MP3.

It is sad that many young physicists at this level cannot negotiate their way correctly through a straightforward gas-law calculation. As usual, many failed to convert from degree Celsius to Kelvin before carrying out the numerical manipulation. This was an excellent way to lose marks.

A good number were able to give answers to parts of this question but few could pull all the strands together convincingly. Frequent omissions were: to show that the internal energy changes are identical because the endpoint temperatures are the same, and to use the first law to confirm the final linking statement.