Question

An ideal monatomic gas expands adiabatically from an initial temperature of 384 K and volume of 4.6 m³ to a final volume of 10 m³. If the initial pressure is 1.5 ? P_atm, how much work is done on the gas?

Answer 1

In an adibatic process no heat is transferred. Therefore the work done on the gas equals its change in internal energy:
W = ?U

The change in internal energy for an ideal gas is given by
?U = n?Cv??T = n?Cv?(T? - T?)
(subscripts 2 denotes final state 1 initial state)
The molar heat capacity at constant volume of a monatomic ideal gas is:
Cv = (3/2)?R
So
?U = (3/2)?n?R?(T? - T?)
Form ideal gas law follows
n?R?T = P?V
So
?U = (3/2)?(P??V? - P??V?)
= (3/2)?P??V??( (P?/P?)?(V?/V?)? - 1)

Assuming reversible operation, pressure and volume for an ideal gas undergoing an adiabatic change of state are related as:
P?V^? = constant
? is the heat capacity ratio, For a monatomic ideal gas:
? = Cp/Cv = (5/2)?R / (3/2)?R = 5/3
So pressure in initial and final state are related as:
P??V?^? = P??V?^?
<=>
(P?/P?) = (V?/V?)^(-?)

Hence,
W = ?U = (3/2)?P??V??( (V?/V?)^(1-?)? - 1)
= (3/2) ? 1.5?101325 Pa ? 4.6 m