Question

A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.

Answer 1

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

Best regards!