Question

A gas at 155 kPa and standard temperature has an initial volume of 1.00 L. The pressure of the gas rises to 500 kPa as the temperature also rises to 135°C. What is the new volume?

2.16 L
0.463 L
0.207 L
4.82 L

Answer 1

In the combined gas equation, (P1V1)/T1=(P2V2)T2, where P1 is the initial pressure, V1 the initial volume, T1 the initial temperature,P2 is the final pressure, V2 the final volume, T2 the final temperature,

substituting for the values: (155kPa× 1.00L)/273K=(500kPA× V2)/408K

V2=(155kPa×1.00L×408K)/(273K×500kPa)

= 0.463L