Answer to Question #302459 in General Chemistry for Loni

Question

The initial temperature of a 1.00 liter sample of argon is 20 C. The pressure is decreased from 720 mmHg to 360 mmHg and the volume increases to 2.14 liters. What was the change in temperature of the argon?

Answer 1

V1 = 1 L

T1 = 20°C = 293°K

P1 = 720 mHg = 0.947369 atm

P2 = 360 mmHg = 0.473684 atm

V2 = 2.14 L

Insert all known values into the following formula:

P1V1/T1 = P2V2/T2

0.947369 x 1/293 = 0.473684 x 2.14/T2

Solving for T2 = 313.51°K or 40.51°C.

So, temperature change from 20°C to 40.51°C

= 20.51°C