Answer to Question #167067 in General Chemistry for Nada

Question

A 9.40 L

 container holds a mixture of two gases at 43 °C.

 The partial pressures of gas A and gas B, respectively, are 0.321 atm

 and 0.658 atm.

 If 0.120 mol

 of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

partial pressures of gas A, P_A = 0.321 atm

partial pressures of gas B, P_B = 0.658 atm

Total pressure, P = P_A+P_B

= (0.321+0.658) atm

= 0.979 atm

Volume of the container, V = 9.40 L

Temperature,T = (43+273)K

= 316 K

Let, Number of moles of the gas = n

According to the ideal gas equation,

PV =nRT

(R = 0.082 L.atm.K-1.mol-1)

Or, n =(PV/RT)

Or, n = [(0.979×9.40)/(0.082×316)]

Or, n = 0.355

After adding 0.120 mol of a third gas total number of moles,

n' = (0.355+0.120)

= 0.475 mol

Keeping the volume,V = 9.4 L

And temperature, T = 316 K

So, from ideal gas equation,

P =(n'RT/V)

Or, P = (0.475×0.082× 373/9.4)

= 1.546 atm

So, pressure of the gas after adding 0.120 mol of 3rd gas = 1.546 atm