# A 5 L container holds 16  mol and 6  mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from 320^oK to 450 ^oK. By how much does the pressure change?

Apr 21, 2016

56.4% increase

#### Explanation:

Caution : Dont write ${X}^{o}$ if it is in kelvin, write $X K$ straightaway.

Considering they are ideal gases, they follow

$P V = n R T$

Here, $V$ is constant and $R$ is anyway constant, so we are left with

${P}_{1} / \left({n}_{1} {T}_{1}\right) = {P}_{2} / \left({n}_{2} {T}_{2}\right)$

Note that, $\frac{16}{6} > \frac{4}{3}$. So gas B gets totally consumed up. So, $\frac{6}{3} = 2$ moles of new has was formed, so in the process $8$ moles of gas A was taken up and $8$ moles of A is remaining. So in the reaction, total ${n}_{1} \left(= 16 + 6 = 22\right)$ moles of gas is transformed into ${n}_{2} \left(= 2 + 8 = 10\right)$ moles of gas.

we are given,

${T}_{1} = 320 K$
${T}_{2} = 450 K$

So, ${P}_{1} / {P}_{2} = \frac{{n}_{1} {T}_{1}}{{n}_{2} {T}_{2}} = \frac{22 \cdot 320}{10 \cdot 450} \approx 1.564$

So pressure is about 56.4% increased.