A gas originally occupying 10.1 L at 0.925 atm and 25°C is changed to 13.3 L at 625 torr. What is the new temperature?
A) 349 °C
B) 76 °C
C) -244 °C
D) 622 °C
E) 50 °C

Question

Quizplus · Accepted Answer

Using the combined gas law formula:

(P1V1/T1) = (P2V2/T2)

Converting the initial pressure to torr:

0.925 atm x 760 torr/atm = 702.8 torr

Plugging in the values:

(702.8 torr)(10.1 L)/(298 K) = (625 torr)(13.3 L)/T2

Solving for T2:

T2 = (625 torr)(13.3 L)(298 K)/(702.8 torr)(10.1 L)

T2 = 349 K

Converting to Celsius:

T2 = 349 K - 273 = 76 °C

Therefore, the answer is B.

A Gas Originally Occupying 10