Physics 213 Problem 1 Week 3
Adiabatic Compression
a) Last week, we considered the problem of isothermal compression: 1.5 moles of an ideal
diatomic gas at temperature 35oC were compressed isothermally from a volume of 0.015 m3
to a volume of 0.0015 m3. The pV-diagram for the isothermal process is shown below. Now
we consider an adiabatic process, with the same starting conditions and the same final
volume. Is the final temperature higher, lower, or the same as the in isothermal case? Sketch
the adiabatic processes on the p-V diagram below and compute the final temperature. (Ignore
vibrations of the molecules.)
For an adiabatic process ViTiα = VfTfα , where α = 5/2 for the diatomic gas. In this case Ti = 273
K + 35oC = 308 K. We find Tf = Ti (Vi/Vf)1/α = (308 K) (10)2/5 = 774 K.
b) According to your diagram, is the final pressure greater, lesser, or the same as in the
isothermal case? Explain why (i.e., what is the energy flow in each case?). Calculate the
final pressure.
We argued above that the final temperature is greater for the adiabatic process. Recall that p = nRT/ V
for an ideal gas. We are given that the final volume is the same for the two processes. Since the final
temperature is greater for the adiabatic process, the final pressure is also greater. We can do the
problem numerically if we assume an idea gas with constant α.
In an adiabatic processes, pVγ = constant, where γ = (α + 1) / α. Here, the gas is diatomic and
non-vibrational, so α = 5/2, and γ = 7/5. Note that γ is always greater than one. Then for an
adiabatic process, pfVf
γ = piVi
γ , or pf = pi (Vi/Vf) γ. For an isothermal process, pfVf = piVi. or pf =
pi (Vi/Vf). Since γ > 1, the final pressure must be greater in the adiabatic process. Numerically,
pf = pi (Vi/Vf) γ = 2.56 × 105 Pa × (0.015 m3/0.0015 m3)7/5 = 6.43 × 106 Pa
Recall that in the isothermal process the final presume was 2.56 × 105 Pa. Thus the final
pressure is greater in the adiabatic process.
Solution
