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?
1 Answer
The pressure changed by:
Explanation:
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: