Solution
Physics 213 Problem 4 Week 2
Isothermal Compression
a) One and one-half moles of a diatomic gas at temperature 35oC are compressed isothermally
from a volume of 0.015 m3 to a volume of 0.0015 m3. Sketch the process on a pV diagram
and show what corresponds to the work done on the gas. Calculate the work done on the gas.
(Ignore vibrational energy of the molecules.)
The figure above shows the path along which the gas is compressed from the initial state to the
final state. The path is an isotherm. The equation of state of an ideal gas is p = nRT/V. The
differential of the work done by the gas is dW = pdV. Substituting for p, we obtain dW =
(nRT/V) dV. Since the temperature, T, is constant (isothermal process), we can integrate the
work from the initial to the final volume. Note that the work done by the gas is the negative of
the work done on the gas: Wby = − Won.
Won =−Wby =− pdV
Vi
Vf
∫=−nRT dV
V=nRTln Vi
Vf
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
Vi
Vf
∫
Won = = (1.5 mol) (8.314 J / mol·K) (273 K + 35 K) ln(0.015 m3/ 0.0015 m3) = 8844 J
Positive work is done on the gas.
Note that we did the integral for Wby by substituting nRT/V for p and took the factor nRT out of
the integral sign. This procedure is correct only for isothermal (T = constant) processes. Had
the process been, for example, isobaric (constant pressure), we would have simply brought p out
of the integral and found that Wby = p (Vf − Vi). The point is that the way we do the integral
varies with the nature of the process. Wby = nRT ln(Vf / Vi) for isothermal processes only.
