| Date | May 2016 | Marks available | 6 | Reference code | 16M.3.SL.TZ0.8 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Calculate, Determine, and Draw | Question number | 8 | Adapted from | N/A |
Question
A fixed mass of an ideal monatomic gas undergoes an isothermal change from A to B as shown.
The temperature at A is 350 K. An identical mass of the same ideal monatomic gas undergoes an isobaric change from A to C.
(i) Calculate the temperature at C.
(ii) Calculate the change in internal energy for AC.
(iii) Determine the energy supplied to the gas during the change AC.
(iv) On the graph, draw a line to represent an adiabatic expansion from A to a state of volume 4.0×10−3m3 (point D).
(i) State the change in entropy of a gas for the adiabatic expansion from A to D.
(ii) Explain, with reference to the concept of disorder, why the entropy of the gas is greater at C than B.
Markscheme
(i) 1400 «K»
(ii) \(\frac{3}{2}P\Delta V = \frac{3}{2} \times 4 \times {10^5} \times 3 \times {10^{ - 3}}\)
1800 J
(iii) 1800+PΔV=1800+4×105×3×10−3 OR use of \(\Delta Q = \frac{5}{2}P\Delta V\)
3000 J
(iv) curve starting at A ending on line CB AND between B and zero pressure
(i) 0
(ii)
ALTERNATIVE 1
C has the same volume as B OR entropy is related to disorder
higher temperature/pressure means greater disorder
therefore entropy at C is greater «because entropy is related to disorder»
ALTERNATIVE 2
to change from B to C, ΔQ > 0
so ΔS > 0
ΔS related to disorder
