Question

A 8.70 L 8.70 L container holds a mixture of two gases at 33 ° C. 33 °C. The partial pressures of gas A and gas B, respectively, are 0.314 atm 0.314 atm and 0.755 atm. 0.755 atm. If 0.200 mol 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Answer:

Total pressure is equal to = Summation of all partial pressure = 1.64 atm

Explanation

Since partial pressure of gas A and B is given.

So at first we calculate partial pressure of gas C using ideal gas equation.

Volume of container V = 8.7 lit

Moles of gas C is added = 0.200

Temperature = 33°C = 306 Kelvin

We know PV = nRT

where R is equal to = 0.0821 lit. atm. mole⁻¹ K⁻¹

So Pc = $\frac{nRT}{V}$ = $\frac{0.2 X 0.0821 X 306}{8.7}$ = 0.577 atm

Given Partial pressure of A  = 0.314 atm

and Partial pressure of B = 0.755 atm

So total pressure = 0.577 + 0.314 + 0.755 = 1.64 atm