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QJEEAdv19Pha188: JEE Advanced 2019 Physics - Class 11 JEEAdv19Pha188 Solution: A mixture of ideal gas containing 5 moles of monatomic gas and…

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P0, volume V0, and temperature T0. If the gas mixture is adiabatically compressed to a volume V04{{{V_0}} \over 4} , then the correct statement(s) is/are

(Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)

Multiple Choice Options

  • A The final pressure of the gas mixture after compression is in between 9P0 and 10P0
  • B The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0
  • C Adiabatic constant of the gas mixture is 1.6
  • D The work |W| done during the process is 13RT0

Step-by-Step Verified Solution

γmix=n1Cp1+n2Cp2n1Cv1+n2Cv2=85{\gamma _{mix}} = {{{n_1}{C_{p1}} + {n_2}{C_{p2}}} \over {{n_1}{C_{v1}} + {n_2}{C_{v2}}}} = {8 \over 5}

W=P1V1−P2V2γ−1W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}

P0V08/5=P2(V04)8/5{P_0}{V_0}^{8/5} = {P_2}{\left( {{{{V_0}} \over 4}} \right)^{8/5}}

P2=9.2P0{P_2} = 9.2{P_0}

W=P0V0−9.2P0V043/5=−13RT0W = {{{P_0}{V_0} - 9.2{P_0}{{{V_0}} \over 4}} \over {3/5}} = - 13R{T_0}

∴\therefore ∣W∣=13RT0|W| = 13R{T_0}

(b) T1V1γ−1=T2V2γ−1{T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}

T2=T1(2)6/5=23T0{T_2} = {T_1}{(2)^{6/5}} = 23{T_0}

Average kinetic energy of gas mixture

= nCvmixT2=23RT0nC{v_{mix}}{T_2} = 23R{T_0}

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