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
Aug 1, 2016

Answer:

The pressure changed by: #2.39 × 10^6 Pa# (to 3 sig figs)

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: #pV = nRT#
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:
#p = (nRT)/V#
#⇒ p_1 - p_2 = (n_1RT_1)/V_1 - (n_2RT_2)/V_2#

The question implies that the volume does not change so #V_1 = V_2#
#⇒ p_1 - p_2 = ((n_1RT_1) - (n_2RT_2))/V#

Before calculation the volume needs to be converted into cubic metres: #V = 5L = 5×10^(-3) m³#

Substitute in the values
#⇒ p_1 - p_2 = ((18 × 8.31 × 360) - (12 × 8.31 × 420))/(5×10^(-3)) = 2.393 × 10^6 Pa#

The pressure changed by:
#2.39 × 10^6 Pa# (to 3 sig figs)