Question

A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is and the initial temperature is The final temperature of the gas is What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).

A) 350 L
B) 270 L
C) 190 L
D) 230 L

Answer 1

There are some missing data in the text of the exercise. Here the complete text:
"A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "

Solution:

First, we can find the initial volume of the gas, by using the ideal gas law:
$pV=nRT$
where 
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature

Using the initial data of the gas, we can find its initial volume:
$V_i = \frac{nRT_i}{p_i} = \frac{(20.0 mol)(8.31 J/molK)(450 K)}{4 \cdot 10^5 Pa} =0.187 m^3$

Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
$TV^{\gamma-1} = cost.$
where $\gamma=1.67$ for a monoatomic gas as in this exercise. The previous relationship can be also written as
$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$
where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
$V_f = V_i \sqrt[\gamma-1]{ \frac{T_i}{T_f} }=(0.187 m^3) \sqrt[0.67]{ \frac{450 K}{320 K} }=0.310 m^3 = 310 L$
So, the final volume of the gas is 310 L.