Question

A mixture of xenon and hydrogen gases, in a 9.54 L flask at 32 °C, contains 21.0 grams of xenon and 0.792 grams of hydrogen. The partial pressure of hydrogen in the flask is ___ atm ans the total pressure in the flask is ___ atm.

Answer 1

The Xenon and hydrogen gases are considered as ideal gases. They are non reactive with each other. Thus, we will use ideal gas equation to determine the total pressure exerted on the system.

Given, weight of xenon = 21g

weight of hydrogen = 0.792g

We need to convert them into moles.

we know, moles = given weight / molecular weight

M( hydrogen gas ) = 2g/mol

M ( Xenon gas ) = 131 g/mol

n( H2 ) = 0.792g / 2g mol^-1 = 0.396mol

n(Xe) = 21g / 131g mol^-1 = 0.1603mol

Total no. of moles present in the container = 0.396mol + 0.1603 mol= 0.5563 mol

The ideal gas equation is

PV = nRT

Given, V = 9.54L

T = 32 + 273K = 305K

n = 0.5563 mol

R = ideal gas constant = 0.08205 L atm / mol K

so, P = nRT / V =

The total pressure (P) exerted by both gases is 1.462atm.

According to Daltons's law of partial pressure, the total pressure is the sum of partial pressure of individual gas present in the system.

Hence, p(hydrogen) = P * X_H

Where,p(Hydrogen) = partial pressure of the hydrogen gas

X_H is the mole fraction of hydrogen gas present.

we know, mole fraction = No. of moles of a component / total no. of moles present in the system.

Hence, X_H = 0.396 mol / 0.5563 mol = 0.712

p(hydrogen) = 1.462 atm * 0.712 = 1.04 atm