Question

A 9.40-L container holds a mixture of two gases at 45 °C. The partial pressures of gas A and gas B, respectively, are 0.181 atm and 0.780 atm. If 0.100 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

V= 9.40 L

T= 45 oC = 318.15 K

PA= 0.181 atm

PB = 0.780

PT1= 0.181+0.780= 0.961 atm

Moles of third gases added= 0.100 mol.

R= 0.082 L atm K⁻¹ mol⁻¹

Assume that these gases are ideal gases

Hence, PV=n₁RT

n₁ = PV/RT= (0.961 * 9.40)/(0.082 * 318.15)

n₁ = 0.3462 mol

Now 0.100 mol of third gas is added to the system, volume & temperature constant

n₂ = n₁ +0.100 = 0.3462+0.100 = 0.4462 mol

then P_T2 V = n₂RT

   P_{T2 =} n₂RT/V = (0.4462 * 0.082 *318.15)/9.40 = 1.238 atm

the total pressure become after a third gas is added with no change in volume or temperature is 1.238 atm