JEE Advance - Physics (2019 - Paper 2 Offline - No. 8)
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 $${{{V_0}} \over 4}$$, then the correct statement(s) is/are
(Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)
(Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)
The final pressure of the gas mixture after compression is in between 9P0 and 10P0
The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0
Adiabatic constant of the gas mixture is 1.6
The work |W| done during the process is 13RT0
คำอธิบาย
$${\gamma _{mix}} = {{{n_1}{C_{p1}} + {n_2}{C_{p2}}} \over {{n_1}{C_{v1}} + {n_2}{C_{v2}}}} = {8 \over 5}$$
$$W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$$
$${P_0}{V_0}^{8/5} = {P_2}{\left( {{{{V_0}} \over 4}} \right)^{8/5}}$$
$${P_2} = 9.2{P_0}$$
$$W = {{{P_0}{V_0} - 9.2{P_0}{{{V_0}} \over 4}} \over {3/5}} = - 13R{T_0}$$
$$ \therefore $$ $$|W| = 13R{T_0}$$
(b) $${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$
$${T_2} = {T_1}{(2)^{6/5}} = 23{T_0}$$
Average kinetic energy of gas mixture
= $$nC{v_{mix}}{T_2} = 23R{T_0}$$
$$W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$$
$${P_0}{V_0}^{8/5} = {P_2}{\left( {{{{V_0}} \over 4}} \right)^{8/5}}$$
$${P_2} = 9.2{P_0}$$
$$W = {{{P_0}{V_0} - 9.2{P_0}{{{V_0}} \over 4}} \over {3/5}} = - 13R{T_0}$$
$$ \therefore $$ $$|W| = 13R{T_0}$$
(b) $${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$
$${T_2} = {T_1}{(2)^{6/5}} = 23{T_0}$$
Average kinetic energy of gas mixture
= $$nC{v_{mix}}{T_2} = 23R{T_0}$$