A #5 L# container holds #12 # mol and #6 # mol of gasses A and B, respectively. Every four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from #360^oK# to #420 ^oK#. By how much does the pressure change?
To get started we’ll work out how many moles of gas are present at the end of the reaction, we start with 12 moles of gas A and 6 moles of gas B (a total of 18 moles). In the reaction gas A bonds with gas B at a ratio of 3:4 molecules, so that ratio will apply to the number of moles also.
There are only 6 moles of gas B which is not a multiple of 4, so only 4 moles of gas B will react and only 3 moles of gas A will react (due to 3:4 ratio). That gives only 1 mole of the new gas from the reaction. In the end we have 9 moles of gas A left, 2 moles of gas B and 1 mole of the new gas, a total of 12 moles.
Ideal Gas Equation
Use the ideal gas equation to find the pressure change:
p is pressure (Pa)
V is volume (m³)
n is number of moles (mol)
R is the molar gas constant (8.31 m² kg s⁻² K⁻¹ mol⁻¹)
T is temperature (K)
Here is the equation with pressure as the subject:
The question implies that the volume does not change so
Before calculation the volume needs to be converted into cubic metres:
Substitute in the values
The pressure changed by: