# 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:*