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

`P_A=4.20atm\\\\P_B=17.1atm`

Step-by-step 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=(n_ART)/(V) =(1.21mol*0.082(atm*L)/(mol*K)*301.25K)/(7.12L) \\\\P_A=4.20atm\\\\P_B=(n_BRT)/(V) =(4.94mol*0.082(atm*L)/(mol*K)*301.25K)/(7.12L) \\\\P_B=17.1atm`

Best regards!