Question

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?

Answer 1

`\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%`