Submit At 25.0 C, a 10.00 L vessel is filled with 5.00 atm of Gas A and 7.89 atm of Gas B. What is the mole fraction of Gas B?

the mole fraction of Gas B is xB = 0.612 (61.2%)

Explanation:

Assuming ideal gas behaviour of A and B, then

pA*V=nA*R*T

pB*V=nB*R*T

where

V = volume = 10 L

T = temperature = 25°C = 298 K

pA and pB = partial pressures of A and B respectively = 5 atm and 7.89 atm

R = ideal gas constant = 0.082 atm*L / (mol*K)

therefore

nA = (pA*V) / (R*T) = 5 atm * 10 L / (0.082 atm*L / (mol*K) * 298 K) = 2.04 mole

nB = (pB*V) / (R*T) = 7.89 atm * 10 L / (0.082 atm*L / (mol*K) * 298 K) = 3.22 mole

therefore the total number of moles is

n = nA + nB = 2.04 mole + 3.22 mole = 5.26 mole

the mole fraction of Gas B is then

xB = nB/n = 3.22 mole/5.26 mole = 0.612

xB = 0.612

Note

another way to obtain it is through Dalton's law

P=pB*xB, P = pA+pB → xB = pB / (pA+pB) = 7.69 atm / (5 atm + 7.89 atm) = 0.612