Question

A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What is the final pressure in atm?

Answer 1

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm