Question

According to the ideal gas law, a 10.68 mol sample of methane gas in a 0.8295 L container at 501.9 K should exert a pressure of 530.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure

Answer 1

The answer is "152.28%".

Step-by-step explanation:

Formula for ideal gas:

PV=nRT

equation:

`\to (P+(an^2)/(v^2))(v-nb)=nRT \\\\a=2.253 (L^2 atm)/(mol^2)\\\\b=4.278 * 10^(-2) (L)/(mol)\\\\n=10.68 \ mol\\\\v= 0.8295 \ L\\\\\to (P+(an^2)/(v^2))= (nRT)/((v-nb))\\\\\to (P)= (nRT)/((v-nb)) - (an^2)/(v^2) \\\\`

`= (10.68 * 0.0082 * 501.9)/(0.8295 - (10.68 * 4.278 * 10^(-2))) - (2.253 * 10.68^2)/(0.8295^2)\\\\=(43.9543944)/(0.4568904) - (256.982587)/(0.68807025)\\\\ = 96.2033661- 373.483067\\\\=-277.279701 \\\\ =-277.279701 \ atm`

The pressure of gas:

`= (530.3 + 277.279701) \ atm\\\\= 807.579701 \ atm \\\\`

Calculating the pressure percentage:

`=(807.579701)/(530.3) * 100\\\\= 152.28 \%`