Question

A gas at a pressure of $5.0$ bar is heated from $0^o$ to ${546}^{o}C$ and is simultaneously compressed to one-third of its original volume. Hence, final pressure is:

• A. $\frac{5}{9}$bar
• B. $15.0$bar
• C. $30.0$bar
• D. $45.0$bar

Answer 1

The correct option is D $45.0$bar

Initial volume of the gas$=V_{1}L$

Initial Temperature of the gas$=T_1=(0+273)=273K$

Final pressure of the gas$=P_2=?$

Final volume of the gas$=V_{2}L$

Final Temperature of the gas$=T_2=(546+273)K=819K$

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

so $V_1=3V_2$

We know from gas laws,

$\frac{P_2{V}_2}{819}=\frac{5\times 3V_2}{273}$

$P_2=\frac{5\times 3\times 819}{273}bar=45bar$