# A 5 L container holds 8  mol and 5  mol of gasses A and B, respectively. Groups of five molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 100 K to 240 K. How much does the pressure change?

Dec 26, 2017

The reaction decreases the number of moles of gas in the (constant volume) container from $13$ to $7$ while increasing the temperature from $100$ $K$ to $240$ $K$.

The pressure change is $2794 - 2162 = 632$ $k P a$

#### Explanation:

First we need to write a balanced chemical equation for the reaction:

$2 A + 5 B \to {A}_{2} {B}_{5}$

Since we have $5$ mol of $B$, all of it will be used up in the reaction. Only $2$ of the $8$ mol of $A$ will be used, so $6$ mol will remain.

After the reaction, therefore, there will be $6$ mol of $A$ and $1$ mol of $B$ for a total of $7$ mol. There were initially $13$ mol of gas in the container.

We remember that a mole of any gas that behaves like an ideal gas takes up the same volume under the same temperature and pressure conditions.

Let's call the pressure before the reaction ${P}_{1}$ and after the reaction the pressure is ${P}_{2}$.

$P V = n R T$ where $R$ is the gas constant, $8.314$ $L k P a {K}^{-} 1 m o {l}^{-} 1$

${P}_{1} = \frac{n R T}{V} = \frac{13 \times 8.314 \times 100}{5} = 2162$ $k P a$

${P}_{2} = \frac{n R T}{V} = \frac{7 \times 8.314 \times 240}{5} = 2794$ $k P a$

The pressure change is $2794 - 2162 = 632$ $k P a$

Dec 27, 2017

29% increase

#### Explanation:

Idea gas law:

PV=nRT

For mix gases inside the same volume and temperature

PV$= \left({n}_{A} + {n}_{B}\right) R T$

Let $\left({n}_{1} , {P}_{1} , {V}_{1} , {T}_{1}\right) \mathmr{and} \left({n}_{2} , {P}_{2} , {V}_{2} , {T}_{1}\right)$ denote the number of moles, pressure, volume, and temperature before and after the reaction happened respectively.

Before the reaction, there are

${n}_{1} = {n}_{A} + {n}_{B} = 8 + 5 = 13 m o l$ of mixed gases

The reaction

$\cancel{5 B \to 2 A}$ (This reaction is incorrect, I misread the question, B binds with A, not binds to form A)

$5 B + 2 A \to {A}_{2} {B}_{5}$

Takes away 5 moles of B and 2 moles of A into one moles of ${A}_{2} {B}_{5}$, leaving 6 mol of A.

${n}_{2} = {n}_{A} + {n}_{{A}_{2} {B}_{5}}$ = 6+1 = 7 mol of mixed gases.

The idea gas law before and after the reaction are:

${P}_{1} {V}_{1} = {n}_{1} R {T}_{1}$

${P}_{2} {V}_{2} = {n}_{2} R {T}_{2}$

$\frac{{P}_{2} {V}_{2}}{{P}_{1} {V}_{1}} = {n}_{2} / {n}_{1} \frac{R {T}_{2}}{R {T}_{1}}$

For a fixed container, ${V}_{1} = {V}_{2}$

${P}_{2} / \left({P}_{1}\right) = \frac{7}{13} \cdot \frac{240 K}{100 K} = 1.29$

The change in pressure is

(P_2-P_1)/p_1 = (DeltaP)/P_1 =P_2/P_1 -1 = 0.29 or 29%

The pressure has thus increased 85%.

Since the volume of the container is given, ${P}_{1}$ can be reality calculated to determine the absolute change in pressure ($\Delta P$)

P_1 = (n_1 RT_1)/V = ((13mol*8.3145 J/(mol K)*100K)/(5L*0.001m^3/L)) /(1.01325 "x" 10^5 N/(m^2atm)) ~ 21.34 atm

DeltaP ~0.29P_1= 0.29* 21.3 atm = 6.2 atm