Question
Pressure \(p\), volume \(V\) and temperature \(T\) are measured for a fixed mass of gas. \(T\) is measured in degrees Celsius.
The graph shows the variation of \(p V\) with \(T\).
The mass of a molecule of the gas is \(4.7 \times 10^{-26} \mathrm{~kg}\).
(a) State the unit for pV in fundamental SI units. [1]
(b) Deduce, using the graph, whether the gas acts as an ideal gas. [3]
(c) Calculate, in g, the mass of the gas. [3]
▶️Answer/Explanation
Ans:
a.\(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\)
b.ALTERNATIVE 1
Graph shown is a straight line/linear
OR
expected graph should be a straight line/linear \(\checkmark\)
If ideal then \(T\) intercept must be at \(T=-273^{\circ} \mathrm{C}\)
Use of \(y=m x+c\) to show that \(x=-273^{\circ} \mathrm{C}\) when \(y=0\)
(hence ideal)
ALTERNATIVE 2
Calculates \(\frac{p V}{T}\) for two different points
Obtains \(1.50 \ll \mathrm{J} \mathrm{K}^{-1}\) for both
States that for ideal gas \(\frac{p V}{T}=n R\) which is constant and concludes that gas is ideal
c.Use of \(n=\frac{p V}{R T}\) OR \(N=\frac{p V}{k T}\)
Mass of gas \(=n \times N_{\mathrm{A}} \times\) mass of molecule
OR
Mass of gas \(=N \times\) mass of molecule
\(5.1 \propto g n \checkmark\)
This question is in two parts. Part 1 is about ideal gases and specific heat capacity. Part 2 is about simple harmonic motion and waves.
Part 1 Ideal gases and specific heat capacity
a
State two assumptions of the kinetic model of an ideal gas.
Argon behaves as an ideal gas for a large range of temperatures and pressures. One mole of argon is confined in a cylinder by a freely moving piston.
(i) Define what is meant by the term one mole of argon.
(ii) The temperature of the argon is 300 K. The piston is fixed and the argon is heated at constant volume such that its internal energy increases by 620 J. The temperature of the argon is now 350 K.
Determine the specific heat capacity of argon in J kg–1 K–1 under the condition of constant volume. (The molecular weight of argon is 40)
At the temperature of 350 K, the piston in (b) is now freed and the argon expands until its temperature reaches 300 K.
Explain, in terms of the molecular model of an ideal gas, why the temperature of argon decreases on expansion.
Answer/Explanation
Markscheme
a
point molecules / negligible volume;
no forces between molecules except during contact;
motion/distribution is random;
elastic collisions / no energy lost;
obey Newton’s laws of motion;
collision in zero time;
gravity is ignored;
(i) the molecular weight of argon in grams / 6.02×1023 argon
atoms / same number of particles as in 12 g of C-12;
(allow atoms or molecules for particles)
(ii) mass of gas = 0.040kg ;
specific heat = \(\frac{Q}{{m\Delta T}}\) or 620 = 0.04×c×50;
(i.e. correctly aligns substitution with equation)
\( = \left( {\frac{{620}}{{0.040 \times 50}} = } \right)310{\rm{Jk}}{{\rm{g}}^{{\rm{ – 1}}}}{{\rm{K}}^{{\rm{ – 1}}}}\);
c.
temperature is a measure of the average kinetic energy of the molecules;
(must see “average kinetic” for the mark)
energy/momentum to move piston is provided by energy/momentum of molecules that collide with it;
the (average) kinetic energy of the gas therefore decreases;
Do not allow arguments in terms of loss of speed as a result of collision with a moving piston.
Examiners report
a
Many could only give one sensible assumption of the ideal gas kinetic model. It was very common to see the bald statement that there are no interatomic forces between the molecules. Candidates failed to give the proviso that this is not true during the collisions between molecules and with the walls of the container. Some candidates tried unsuccessfully to convince examiners that the empirical gas laws are in themselves assumptions.
(i) Many could either define the mole of argon in terms of 12 g of carbon-12 or in terms of a correctly stated Avogadro number. Either was acceptable if clear.
(ii) Although almost all were able to identify the starting point for the calculation of the specific heat capacity of argon, a very common error was to forget that the molar mass is quoted in grams not kilograms. It was therefore common to see answers that were 1000 times too small.
Explanations for the decrease in temperature of the gas on expansion were weak. The key to the explanation is that, at the molecular level, temperature is a measure of the average kinetic energy of the particles. This was often missing from the answers..
An ideal monatomic gas is kept in a container of volume 2.1 × 10–4 m3, temperature 310 K and pressure 5.3 × 105 Pa.
The volume of the gas in (a) is increased to 6.8 × 10–4 m3 at constant temperature.
a.i.
State what is meant by an ideal gas.
Calculate the number of atoms in the gas.
Calculate, in J, the internal energy of the gas.
Calculate, in Pa, the new pressure of the gas.
Explain, in terms of molecular motion, this change in pressure.
Answer/Explanation
Markscheme
a.i.
a gas in which there are no intermolecular forces
OR
a gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures
OR
molecules have zero PE/only KE
Accept atoms/particles.
[1 mark]
N = «\(\frac{{pV}}{{kT}} = \frac{{5.3 \times {{10}^5} \times 2.1 \times {{10}^{ – 4}}}}{{1.38 \times {{10}^{ – 23}} \times 310}}\)» 2.6 × 1022
[1 mark]
«For one atom U = \(\frac{3}{2}\)kT» \(\frac{3}{2}\) × 1.38 × 10–23 × 310 / 6.4 × 10–21 «J»
U = «2.6 × 1022 × \(\frac{3}{2}\) × 1.38 × 10–23 × 310» 170 «J»
Allow ECF from (a)(ii)
Award [2] for a bald correct answer
Allow use of U = \(\frac{3}{2}\)pV
[2 marks]
p2 = «5.3 × 105 × \(\frac{{2.1 \times {{10}^{ – 4}}}}{{6.8 \times {{10}^{ – 4}}}}\)» 1.6 × 105 «Pa»
[1 mark]
«volume has increased and» average velocity/KE remains unchanged
«so» molecules collide with the walls less frequently/longer time between collisions with the walls
«hence» rate of change of momentum at wall has decreased
«and so pressure has decreased»
The idea of average must be included
Decrease in number of collisions is not sufficient for MP2. Time must be included.
Accept atoms/particles.
[2 marks]