Question

A 5.06 L cylinder contains 1.92 mol of gas A and 3.25 mol of gas B, at a temperature of 35.0 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior. КРа Ра 3 КРа Рв —

Answer 1

1) for Gas A:

Given:

V = 5.06 L

n = 1.92 mol

T = 35.0 oC

= (35.0+273) K

= 308 K

use:

P * V = n*R*T

P * 5.06 L = 1.92 mol* 0.08206 atm.L/mol.K * 308 K

P = 9.5903 atm

= 9.5903*101.33 kPa

= 972 kPa

Answer: 972 kPa

2) for Gas B:

Given:

V = 5.06 L

n = 3.25 mol

T = 35.0 oC

= (35.0+273) K

= 308 K

use:

P * V = n*R*T

P * 5.06 L = 3.25 mol* 0.08206 atm.L/mol.K * 308 K

P = 16.2336 atm

= 16.2336*101.33 kPa

= 1645 kPa

Answer: 1645 kPa