CAIE A-Level Physics A2 15 Ideal Gases Questions

Practise calculating amount of substance and particles, applying pV = nRT or pV = NkT and analysing molecular assumptions, pressure and kinetic energy.

Exam points

calculate amount of substance and particle number using the mole and Avogadro constant
apply ideal-gas equations pV = nRT and pV = NkT, including the Boltzmann relationship
analyse molecular assumptions, pressure, rms speed and kinetic energy using kinetic theory

Covered syllabus points

15.1 The mole 15.2 Equation of state 15.3 Kinetic theory of gases

[Maximum number: 5]

A fixed mass of an ideal gas has volume $210 \mathrm{~cm}^{3}$ at pressure $3.0 \times 10^{5} \mathrm{~Pa}$ and temperature 270 K . The volume of the gas is reduced at constant pressure to $140 \mathrm{~cm}^{3}$ , as shown in Fig. 2.1.

Fig. 2.1

The final temperature of the gas is T.

(a)

Determine:

[ 5 ]

(i)

the amount of gas
amount = mol

[ 3 ]

p V=n R T
$n=\left(3.0 \times 10^{5} \times 210 \times 10^{-6}\right) /(8.31 \times 270)$ $=0.028 \mathrm{~mol}$

(ii)

the final temperature T of the gas
T= K

[ 2 ]

$V \propto T$ or T=p V / n R with value of n from (i)
$\begin{aligned} \begin{aligned} T=(140 / 210) \times 270 \text { or } T=\left(3.0 \times 10^{5} \times 140 \times 10^{-6}\right) /(8.31 \times 0.028) =180 \mathrm{~K} \end{aligned} \end{aligned}$

[Maximum number: 8]

(a)

State what is meant by an ideal gas.

[ 2 ]

gas for which $p V \propto T$
where T is thermodynamic temperature

(b)

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.

[ 2 ]

(c)

A sample of $0.26 \mathrm{~m}^{3}$ of an ideal gas is at pressure $2.0 \times 10^{5} \mathrm{~Pa}$ and temperature 290 K .

Determine:

[ 4 ]

(i)

the number N of molecules of the gas

N =

[ 2 ]

p V=N k T
$\begin{aligned} N =\left(2.0 \times 10^{5} \times 0.26\right) /\left(1.38 \times 10^{-23} \times 290\right) =1.3 \times 10^{25} \end{aligned}$

(ii)

the average translational kinetic energy $E_{\mathrm{K}}$ of one molecule of the gas

E_{\mathrm{K}}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{J}[2]

[ 2 ]

$E_{\mathrm{K}}=(3 / 2) k T$ $\begin{aligned} E_{K} =(3 / 2) \times 1.38 \times 10^{-23} \times 290 =6.0 \times 10^{-21} \mathrm{~J} \end{aligned}$

[Maximum number: 9]

(a)

The kinetic theory of gases is based on a number of assumptions about the molecules of a gas.

State the assumption that is related to the volume of the molecules of the gas.

[ 2 ]

(b)

An ideal gas occupies a volume of $2.40 \times 10^{-2} \mathrm{~m}^{3}$ at a pressure of $4.60 \times 10^{5} \mathrm{~Pa}$ and a temperature of $23^{\circ} \mathrm{C}$ .

[ 6 ]

(i)

Calculate the number of molecules in the gas.

number =

[ 3 ]

p V=N k T
$4.60 \times 10^{5} \times 2.40 \times 10^{-2}=N \times 1.38 \times 10^{-23} \times(273+23)$
or

p V=n R T
$\begin{aligned} 4.60 \times 10^{5} \times 2.40 \times 10^{-2}=n \times 8.31 \times(273+23) n=4.49(\mathrm{~mol}) \begin{aligned} N =n N_{\mathrm{A}} =4.49 \times 6.02 \times 10^{23} \end{aligned} \end{aligned}$ $N=2.7 \times 10^{24}$

(ii)

Each molecule has a diameter of approximately $3 \times 10^{-10} \mathrm{~m}$ .

Estimate the total volume of the gas molecules.

volume =m³

[ 3 ]

volume of one atom $=d^{3}\left(=2.7 \times 10^{-29} \mathrm{~m}^{3}\right)$
volume of all atoms $=2.7 \times 10^{-29} \times 2.7 \times 10^{24}$ $=7 \times 10^{-5} \mathrm{~m}^{3}$
or

volume of one atom $=(4 / 3) \pi r^{3}\left(=1.41 \times 10^{-29} \mathrm{~m}^{3}\right)$
volume of all atoms $=2.7 \times 10^{24} \times 1.41 \times 10^{-29}$ $=4 \times 10^{-5} \mathrm{~m}^{3}$

(c)

By reference to your answer in (b)(ii), suggest why the assumption in (a) is justified.

[ 1 ]

numerical comparison between answer to (b)(ii) and $2.4 \times 10^{-2}\left(\mathrm{~m}^{3}\right)$ showing (b)(ii) is much less than $2.4 \times 10^{-2}\left(\mathrm{~m}^{3}\right)$

[Maximum number: 11]

(a)

State what is meant by an ideal gas.

[ 2 ]

(gas that obeys) $p V \propto T$ (for all values of p, V and T )
where T is thermodynamic temperature

(b)

A sealed vessel contains a mass of 0.0424 kg of an ideal gas at $227^{\circ} \mathrm{C}$ . The pressure of the gas is $1.37 \times 10^{5} \mathrm{~Pa}$ and the volume of the gas is $0.640 \mathrm{~m}^{3}$ .

Calculate:

[ 7 ]

(i)

the number of molecules of the gas in the vessel

[ 3 ]

p V=N k T
$N=\left(1.37 \times 10^{5} \times 0.640\right) /\left(1.38 \times 10^{-23} \times(227+273)\right)$ $=1.27 \times 10^{25}$

(ii)

the mass of one molecule of the gas

mass =kg [1]

[ 1 ]

mass $=0.0424 /\left(1.27 \times 10^{25}\right)$ $=3.34 \times 10^{-27} \mathrm{~kg}$

(iii)

the root-mean-square (r.m.s.) speed v of the molecules of the gas.

\begin{aligned} & v= \\ & \mathrm{ms}^{-1} \end{aligned}

[ 3 ]

$\frac{1}{2} m\left\langle c^{2}\right\rangle=(3 / 2) k T$ $3.34 \times 10^{-27} \times v^{2}=3 \times 1.38 \times 10^{-23} \times 500$ $v=2490 \mathrm{~m} \mathrm{~s}^{-1}$
or

$p V=1 / 3(N m)\left\langle c^{2}\right\rangle$ and N m= mass of gas
$0.0424 \times v^{2}=3 \times 1.37 \times 10^{5} \times 0.640$ $v=2490 \mathrm{~m} \mathrm{~s}^{-1}$

(c)

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

[ 2 ]

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Covered syllabus points

Question 1

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Question 2

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Question 4

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