Question

A gas cylinder holds 0.10mol of O2 at 150?Cand a pressure of 3.0atm. The gas expands adiabatically until the pressure is halved.

What's the final volume?

What's the final temperature?

Answer 1

in adiabatic processes, the relationship between P and V is

P V^g = constant where g is gamma, the ratio of specfic heats; for diatomic gases, this is 1.40, so we have

P1 V1^(1.40) = P2 V2^(1.40)

use the perfect gas law to find the initial volume:

P V = n R T

V = n R T/P = 0.1mol x 8.314 Pa-m^3/mol/K x 423K/(3.3x10^5Pa)
V = 0.0012m³

now, using subscript 1 to refer to values before the expansion we have

3.3x10^5Pa (0.00117m^3)^(1.40) = 1.51x10^5Pa V2^1.40

V2 = 0.0019m^3

use the perfect gas law again:

T = P V/(n R) = 1.51x10^5Pa x 0.0019m^3/(0.1 x 8.314) = 346.7K