An $8 L$ $8 L$ container holds $6$ $6$ mol and $6$ $6$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $320 K$ $320K$ to $450 K$ $450 K$ . How much does the pressure change by?

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Answer 1 · 2017-08-18T01:16:44

see below

Given $:$ $:$ $V = 8 L$ $V= 8L$ ; $n_{1} = 6 + 6 = 12 m o l$ $n_1 = 6+6=12mol$

$n_{2} = 3$ $n_2=3$ mol of $A$ $A$ $+ 3$ $+3$ mol of $A B_{2} = 6$ $AB_2=6$ mol

$T_{1} = 320 K$ $T_1=320K$ ; $T_{2} = 450 K$ $T_2=450K$

$R = 8.314 \frac{J}{m o l \cdot K}$ $R=8.314J/(mol*K)$ ; $DeltaP=?$

Applying Ideal gas equation,

$P_1V=n_1RT_1$

$P_2V=n_2RT_2$

$=>P_2=(n_2T_2P_1)/(n_1T_1)$

$=>P_2-P_1=((n_2T_2)/(n_1T_1)-1)P_1$

$=>P_2-P_1=((n_2T_2)/(n_1T_1)-1)((n_1RT_1)/(V))$

$=>P_2-P_1=((6*450)/(12*320)-1)((12*8.314*320)/(8))$

$=>DeltaP=P_2-P_1=-1184.745J/L$

$=>DeltaP=-1184.745(Nm)/(10^-3m^3)$

$=>DeltaP=-1184745$ $Pa$ $=-11.84745$ bar $=-11.6925$ atm

$(-ve)$ sign indicates that the pressure is reduced in the process.

An 8 L8L8 L container holds 6 66 mol and 6 66 mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 320K320K320K to 450 K450K450 K. How much does the pressure change by?