A 7 L container holds 12 mol and 18 mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from 350^oK to 175 ^oK. By how much does the pressure change?

1 Answer
Mar 9, 2016

\Delta P = P-P_0 = (2.494-12.47)\quad MPa = -9.976\quadMPa

Explanation:

Ideal Gas Equation of State: PV=nRT
R = 8.314\quadJK^{-1}mol^{-1}

Before Reaction:
P_{A0} - Partial pressure of species A;
P_{B0} - Partial pressure of species B;
P_0 = P_{A0}+P_{B0} - Total pressure before reaction;

n_{A0} = 12\quad mols - Number of moles of species A;
n_{B0}=18\quad mols - Number of moles of species B;
T_0=350\quad K - Temperature of the mixture;
V_0=7\quad Lit = 7\times10^{-3} \quad m^3 - Volume of the mixture;

P_{A0}=(n_{A0}/V)RT_0 = 4.988\quad MPa;\quad // 1 \quadMPa=10^6\quad Pa
P_{B0}=(n_{B0}/V)RT_0 = 7.483\quad MPa;
P_0 = P_{A0}+P_{B0} = 12.47\quad MPa

After Reaction:
6\quad mols of A would combine with 18\quad mols of B to form 6\quad mols of AB. The final mixture would be 6\quad mols of A and 6\quad mols of AB.

P_{A} - Partial pressure of species A;
P_{AB} - Partial pressure of the new species AB;
P=P_{A}+P_{AB} - Total pressure after reaction;

n_{A} = 6\quad mols - Number of moles of species A;
n_{AB}=6\quad mols - Number of moles of the new species AB;
T=175\quad K - Temperature of the mixture after reaction;
V=7\quad Lit = 7\times10^{-3} \quad m^3 - Volume of mixture;
P_{A}=(n_{A}/V)RT = 1.247 \quad MPa;
P_{AB}=(n_{AB}/V)RT = 1.247\quad MPa;
P = P_{A}+P_{AB} = 2.494\quad MPa

Change in Pressure :
\Delta P = P-P_0 = (2.494-12.47)\quad MPa = -9.976\quadMPa
So the pressure decreases by 9.976 MPa which in terms of percentage is a decrease by 80%