A gas at pressure of 5.0 bar from 0 to 546°c and simultaneously compressed to one third of its original volume.What will be the final pressure in bar?

Oct 1, 2016

Final pressure =5 bar

Explanation:

$\text{Initial Pressure of the gas"=P_1=5" bar}$

$\text{Initial Volume of the gas} = {V}_{1} L$

$\text{Initial Temperature of the gas} = {T}_{1} = \left(0 + 273\right) K = 273 K$

"Final pressure of the gas"=P_2=?

$\text{Final Volume of the gas} = {V}_{2} L$

$\text{Final Temperature of the gas} = {T}_{2} = \left(546 + 273\right) K = 819 K$

Given condition is the is compressed to one third of its original volume.

So ${V}_{1} = 3 {V}_{2}$

We know from gas law

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

$\frac{{P}_{2} {V}_{2}}{819} = \frac{5 \times 3 {V}_{2}}{273}$

${P}_{2} = \frac{5 \times 3 \times {\cancel{819}}^{3}}{\cancel{273}} \text{ bar"=45" bar}$