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=n_{1}RT

n_{1} = PV/RT=
(0.961 * 9.40)/(0.082 * 318.15)

n_{1} = 0.3462
mol

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

n_{2} = n_{1} +0.100 = 0.3462+0.100 = 0.4462
mol

then
P_{T2} V = n_{2}RT

P_{T2 =} n_{2}RT/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

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