QJEEAdv15Ph36fb: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph36fb Solution: A container of fixed volume has a mixture of one mole of…

The internal energy of one mole of an ideal gas at temperature T is given by $U = {f \over 2}RT$ , where f is the degrees of freedom of the gas molecule. The degress of freedom of the gas molecule. The degrees of freedom for hydrogen (diatomic) and helium (monatomic) gases are f_H2 = 5 and f_He = 3, respectively. Thus, ${U_{{H_2}}} = {5 \over 2}RT$ and ${U_{He}} = {3 \over 2}RT$ . The total internal energy of the gas mixture is

${U_{total}} = {U_{{H_2}}} + {U_{He}} = {5 \over 2}RT + {3 \over 2}RT = 4RT$ .

The mixture contains two moles of the gases. The internal energy per mole of the mixture is ${U_{mix}} = {U_{total}}/2 = 2RT$ .

The specific heat at constant volume is given by ${C_v} = dU/dT$ . Thus, the specific heats at constant volume for helium and the mixture are

${C_{v,He}} = d{U_{He}}/dT = {3 \over 2}R$ , and

${C_{v,mix}} = d{U_{mix}}/dT = 2R$

The specific heats at constant pressure, ${C_p} = {C_v} + R$ , for these gases are

${C_{p,He}} = {C_{v,He}} + R = {5 \over 2}R$ , and

${C_{p,mix}} = {C_{v,mix}} + R = 3R$ .

The ratio of specific heats, $\gamma = {C_p}/{C_v}$ , are ${\gamma _{He}} = 5/3$ and ${\gamma _{mix}} = 3/2$ . The speed of sound, in a gas of molecular mass M, is given by ${v_s} = \sqrt {\gamma RT/M}$ . The molecular mass of the gas mixture is

${M_{mix}} = {{{n_{{H_2}}}{M_{{H_2}}} + {n_{He}}{M_{He}}} \over {{n_{{H_2}}} + {n_{He}}}}$

$= {{(1)(2) + (1)(4)} \over {1 + 1}} = 3$ g/mol,

where n_H2 = 1 and n_He = 1 are the number of moles of hydrogen and helium in the gas mixture. The ratio of the speeds of sound in the gas mixture and helium is

${{{v_{s,mix}}} \over {{v_{s,He}}}} = {{\sqrt {{\gamma _{mix}}RT/{M_{mix}}} } \over {\sqrt {{\gamma _{He}}RT/{M_{He}}} }} = \sqrt {{{{\gamma _{mix}}} \over {{\gamma _{He}}}}{{{M_{He}}} \over {{M_{mix}}}}}$

$= \sqrt {{{(3/2)(4)} \over {(5/3)(3)}}} = \sqrt {{6 \over 5}}$

The rms speed of the atoms/molecules is given by ${v_{rms}} = \sqrt {3RT/M}$ . The ratio of the rms speed of helium atoms to that of hydrogen molecules is

${{{v_{rms,He}}} \over {{v_{rms,{H_2}}}}} = {{\sqrt {3RT/{M_{He}}} } \over {\sqrt {3RT/{M_{{H_2}}}} }} = \sqrt {{{{M_{{H_2}}}} \over {{M_{He}}}}} = \sqrt {{2 \over 4}} = \sqrt {{1 \over 2}}$ .