CIE AS & A Level Physics : 15.3 Kinetic theory of gases – Exam style question – Paper 4

Question

(a) Define specific heat capacity.                                                                                                               [2]

(b) A sealed container of fixed volume V contains N molecules, each of mass m, of an ideal gas at pressure p.

 

             (i) State an expression, in terms of V, N, p and the Boltzmann constant k, for the thermodynamic temperature T of the gas.                                                                                  [1]

 

             (ii) Show that the mean translational kinetic energy \(E_K\) of a molecule of the gas is given by

 

                                          \(E_{k}=\frac{3}{2}kT\)                                                                                                                                                                                 [2]            

(iii) Explain why the internal energy of the gas is equal to the total kinetic energy of the molecules.                                                                                                                                         [2]

(c) The gas in (b) is supplied with thermal energy Q.

(i) Explain, with reference to the first law of thermodynamics, why the increase in internal energy of the gas is Q.                                                                                                                        [2]

(ii) Use the expression in (b)(ii) and the information in (c)(i) to show that the specific heat capacity c of the gas is given by

                                                 \(c=\frac{3k}{2m}\)                                    [2]

(d) The container in (b) is now replaced with one that does not have a fixed volume. Instead, the gas is able to expand, so that the pressure of the gas remains constant as thermal energy is supplied. Suggest, with a reason, how the specific heat capacity of the gas would now compare with the value in (c)(ii).                                                                                                                                    [2                                                     [Total: 13]

Answer/Explanation

Ans

(a) (thermal) energy per unit mass (to cause temperature change) 
          (thermal) energy per unit change in temperature B1

 (b) (i) (T =) pV / Nk

 (b) (ii) (pV =) NkT = ⅓Nm < c2>
                 or
                 pV = NkT and pV = ⅓Nm < c2>
                 leading to ½ m <c2> = (3/2)kT and ½ m <c2> = EK 
 (b) (iii) internal energy = Σ EK (of molecules) + Σ EP (of molecules)
                 or
                 no forces between molecules
                 potential energy of molecules is zero 
(c) (i) increase in internal energy = Q + work done 
               constant volume so no work done 
(c) (ii) c = Q / Nm ΔT 
                   = [N × (3/2)kΔT] / (NmΔT) = 3k /2m 
 (d) (as it expands) gas does work (against the atmosphere/external pressure) B1

Question

 A fixed mass of an ideal gas is at a temperature of 21°C. The pressure of the gas is 2.3 × 105 Pa and its volume is 3.5 × 10–3m3.

     (a) (i) Calculate the number N of molecules in the gas.
                                                                                                          N = ………………………………………………… [2]
           (ii) The mass of one molecule of the gas is 40u.
                  Determine the root-mean-square (r.m.s.) speed of the gas molecules.
                                                                                       r.m.s. speed = ………………………………………… ms–1 [2]

 

    (b) The temperature of the gas is increased by 84°C.
           Calculate the value of the ratio

 

                                       

 

                                                                                                     ratio = ………………………………………………… [2]
                                                                                                                                                                       [Total: 6]

Answer/Explanation

Ans

(a) (i) pV =  NkT = or  pV = nRT = and N = nNA

               \(N=\frac{2.3\times 10^{5}\times 3.5\times 10^{-3}}{1.38\times 10^{-23}\times 294}\)

                = 2.0 × 1023

 (a) (ii) \(pV=\frac{1}{3}Nmc^{2}\)

                \(c^{2}=\frac{3\times 2.3\times 10^{5}\times 3.5\times 10^{-3}}{2.0\times 10^{23}\times 40\times 1.66^{-27}}\)

                = 182 000
                r.m.s. speed = 430 ms–1

                or

                \(\frac{1}{2}mc^{2}=\frac{3}{2}kT\)

                 \(c^{2}=\frac{3\times 1.38\times 10^{-23}\times 294}{40\times 1.66\times 10^{-27}}\)

                = 183 000
                 r.m.s.speed = 430 ms–1

(b)  \(c^{2}=\frac{3\times2.0\times10^{23}\times1.38\times 10^{-23}\times(294+84)}{2.0\times 10^{23}\times 40\times 1.66\times 10^{-27}}\)

            c = 236000

            c= 485

            \(ratio=\left ( =\frac{485}{430} \right )=1.1\)

               OR

              \(v\infty \sqrt{T} \ or\ v^{2}\infty T\)

             \(ratio=\sqrt{\frac{273+21+84}{273+21}} \ or\ \sqrt{\frac{378}{294}}\)

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