Question

Twenty moles of a monatomic ideal gas (γ = 5/3) undergo an adiabatic process. The initial pressure is 400 kPa and the initial temperature is 450 K. The final temperature of the gas is 320 K.

In the situation above, the final volume of the gas, in SI units, is closest to:

0.19

0.35

0.23

0.27

0.31

Answer 1

Pressure and volume of an ideal gas undergoing a revisable adiabatic change of state are related as

P∙V^y = C ( a constant )

When you substitute pressure using ideal gas law you can derive a relation for temperature and volume for such process:

P = n*R*T/V

(n*R"T/V)*V^y = C

T*V^(y-1) = C/(n*R) (also a constant)

So, volume and temperature in initial (1) and final (2) state are related as:

T1*V1^((y-1) = T2*V2^(y-1)

T1^(1/(y-1))*V1 = T2^(1/(y-1))*V2

So final volume is given by:

V2 = V1*(T1/T2)^(1/(y-1))

The temperatures are given. The initial volume can be found from ideal gas law:

V1 = n*R*T1/P1

= 20mol * 8.314 J-mol^-1K^-1*450K / 400*10^3 Pa

= 0.187 m^3

= 187 L

Hence,

V2 = 187L * (450K / 320∙)^(1/(1.67 - 1)) = 311 L = 0.31 m^3