Question 90: Thermodynamics & KTG

Question

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)

Accepted Answer

$${\gamma _{mix}} = {{{n_1}{C_{p1}} + {n_2}{C_{p2}}} \over {{n_1}{C_{v1}} + {n_2}{C_{v2}}}} = {8 \over 5}$$
$$W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma  - 1}}$$
$${P_0}{V_0}^{8/5} = {P_2}{\left( {{{{V_0}} \over 4}} ight)^{8/5}}$$
$${P_2} = 9.2{P_0}$$
$$W = {{{P_0}{V_0} - 9.2{P_0}{{{V_0}} \over 4}} \over {3/5}} =  - 13R{T_0}$$
$$ 	herefore $$ $$|W| = 13R{T_0}$$
(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
= $$nC{v_{mix}}{T_2} = 23R{T_0}$$

Answer

The final pressure of the gas mixture after compression is in between 9P0 and 10P0

Answer

The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0

Answer

Adiabatic constant of the gas mixture is 1.6

Answer

The work |W| done during the process is 13RT0

Thermodynamics & KTG

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