Question

A 8.75 L container holds a mixture of two gases at 39 °C. The partial pressures of gas A and gas B, respectively, are 0.407 atm and 0.560 atm. If 0.180 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

PV =nRT

P=pressure= sum of partial pressure of gas A and gas B

partial pressure of gas A =0.407 atm

partial pressure of gas B = 0.560 atm

total pressure P = 0.407 atm +0.560 atm = 0.967 atm

V= volume = 8.75 L

T= 39 degree celsius = 39+273 K= 312 K

R =0.0821 L atm.K-1 .mol-1

n= number of moles of gas A and gas B

n= PV/RT =[ 0.967 atm * 8.75 L]/[ 0.0821 L atm K-1mol-1* 312 K]

=0.330 moles of gas A and gas B

Now when mole of 0.180 of third gas is added then total number of moles of all three gases will be

0.330 mol + 0.180 mol = 0.51 moles

Volume and temperature does not change

so number of moles =0.51 moles

V=8.75 L ; T =312 K

P= nRT/V =[ 0.51 moles * 0.0821 L.atm.K-1.mol-1* 312 K]/ [8.75 L]= 1.49 atm

total pressure =1.49 atm

Answer =1.49 atm