Question

A 8.30-L container holds a mixture of two gases at 15 °C. The partial pressures of gas A and gas B, respectively, are 0.324 atm and 0.883 atm. If 0.140 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Initial total pressure = sum of partial pressure of A and B = 0.324atm + 0.883 atm = 1.207atm

Now we have to determine the initial moles(n) of gas in the system.

By using equation: PV = nRT

n = PV/RT

P = 1.207atm

V = 8.30L

R(gas constant) = 0.082057 Latm.mol^-1K^-1

T = 15⁰C = (15+273)K = 288K

substitude the values in equation n = PV/RT

n = (1.207atm)(8.30L) / (0.082057 Latm.mol^-1K^-1)(288K)

n = 0.42mol (moles of gas initially in system)

0.140mol of a third gas is added so new moles = 0.42mol + 0.140mol = 0.56mol

now we have to find the new pressure by using equation P =nRT/V

V = 8.30L

R(gas constant) = 0.082057 Latm.mol^-1K^-1

T = 15⁰C = (15+273)K = 288K

n = 0.56mol

substitude the values in equation P =nRT/V

P =(0.56mol)(0.082057 Latm.mol^-1K^-1)(288K) / 8.30L

P = 1.59atm