Question 23M.3.SL.TZ2.7
| Date | May 2023 | Marks available | [Maximum mark: 10] | Reference code | 23M.3.SL.TZ2.7 |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Calculate, Explain, Show that, Sketch | Question number | 7 | Adapted from | N/A |
A frictionless piston traps a fixed mass of an ideal gas. The gas undergoes three thermodynamic processes in a cycle.
The initial conditions of the gas at A are:
volume = 0.330 m3
pressure = 129 kPa
temperature = 27.0 °C
Process AB is an isothermal change, as shown on the pressure volume (pV) diagram, in which the gas expands to three times its initial volume.
Calculate the pressure of the gas at B.
[2]
use of pV = constant ✓
PB = 43 «Kpa» ✓
Award [2] for BCA
The gas now undergoes adiabatic compression BC until it returns to the initial volume. To complete the cycle, the gas returns to A via the isovolumetric process CA.
Sketch, on the pV diagram, the remaining two processes BC and CA that the gas undergoes.
[2]
concave curved line from B to locate C with a higher pressure than A ✓
vertical line joining C to A ✓
Allow ECF from MP1 i.e., award [1] for first process locating C at a lower pressure than A, then vertical line to A.
Arrows on the processes are not needed.
Point C need not be labelled.
Show that the temperature of the gas at C is approximately 350 °C.
[2]
ALTERNATIVE 1
use of = constant «so » ✓
TC = 624 «K» OR TC = 351 «°C» ✓
ALTERNATIVE 2
use of to get either pc = OR pc = 268 «kPa» ✓
« TC = 268 × 300/129 = so »
TC = 624 «K» OR TC = 351 «°C» ✓
Explain why the change of entropy for the gas during the process BC is equal to zero.
[1]
ALTERNATIVE 1
«the process is adiabatic so» ΔQ = 0 ✓
ALTERNATIVE 2
The compression is reversible «so ΔS = 0» ✓
OWTTE
Explain why the work done by the gas during the isothermal expansion AB is less than the work done on the gas during the adiabatic compression BC.
[1]
area under curve AB is less than area under curve BC ✓
Do not allow ECF from part (b)
The quantity of trapped gas is 53.2 mol. Calculate the thermal energy removed from the gas during process CA.
[2]
«W = 0 so» Q = ΔU ✓
«ΔU = × 53.2 × R × (351 – 27) so » ΔU = 2.15 × 105 «J» ✓
Award [2] for BCA