CAIE A-Level Physics 15 Ideal Gases
Practise calculating amount of substance and particles, applying pV = nRT or pV = NkT and analysing molecular assumptions, pressure and kinetic energy.
- Syllabus
- 2028–2030
- Course
- Physics 9702
- Level
- A2
Practise calculating amount of substance and particles, applying pV = nRT or pV = NkT and analysing molecular assumptions, pressure and kinetic energy.
A fixed mass of an ideal gas has volume 210 cm3 at pressure 3.0×105 Pa and temperature 270 K . The volume of the gas is reduced at constant pressure to 140 cm3, as shown in Fig. 2.1.

Fig. 2.1
The final temperature of the gas is T.
Determine:
the amount of gas
amount = mol
p V=n R T
C1
n=(3.0×105×210×10−6)/(8.31×270)
C1
=0.028 mol
A1
the final temperature T of the gas
T= K
V∝T or T=p V / n R with value of n from (i)
C1
T=(140/210)×270 or T=(3.0×105×140×10−6)/(8.31×0.028)=180 K
A1
State what is meant by an ideal gas.
(gas that obeys) pV∝T (for all values of p, V and T )
M1
where T is thermodynamic temperature
A1
A sealed vessel contains a mass of 0.0424 kg of an ideal gas at 227∘C. The pressure of the gas is 1.37×105 Pa and the volume of the gas is 0.640 m3.
Calculate:
the number of molecules of the gas in the vessel
p V=N k T
C1
N=(1.37×105×0.640)/(1.38×10−23×(227+273))
C1
=1.27×1025
A1
the mass of one molecule of the gas
mass =0.0424/(1.27×1025)=3.34×10−27 kg
A1
the root-mean-square (r.m.s.) speed v of the molecules of the gas.
21m⟨c2⟩=(3/2)kT
C1
3.34×10−27×v2=3×1.38×10−23×500
C1
v=2490 m s−1
A1
or
pV=1/3(Nm)⟨c2⟩ and N m= mass of gas
(C1)
0.0424×v2=3×1.37×105×0.640
(C1)
v=2490 m s−1
(A1)
The gas in (b) is now cooled gradually to absolute zero.
On Fig. 2.1, sketch the variation with thermodynamic temperature T of the r.m.s. speed of the molecules of the gas.

Fig. 2.1
sketch: line from (0,0) to (500, v)
B1
line with decreasing positive gradient throughout
B1
State what is meant by an ideal gas.
gas for which pV∝T
M1
where T is thermodynamic temperature
A1
Use one of the basic assumptions of the kinetic theory to explain what can be deduced about the potential energy associated with the random motion of molecules in an ideal gas.
no intermolecular forces
B1
(so) potential energy is zero
B1
A sample of 0.26 m3 of an ideal gas is at pressure 2.0×105 Pa and temperature 290 K .
Determine:
the number N of molecules of the gas
p V=N k T
C1
N=(2.0×105×0.26)/(1.38×10−23×290)=1.3×1025
A1
the average translational kinetic energy EK of one molecule of the gas
EK=(3/2)kT
C1
EK=(3/2)×1.38×10−23×290=6.0×10−21 J
A1