• question_answer The pressure and temperature of $4\text{ }d{{m}^{3}}$ of carbon dioxide gas are doubled. Then the volume of carbon dioxide gas would be : A)  $2\text{ }d{{m}^{3}}$                 B)  $3\text{ }d{{m}^{3}}$ C)  $4\text{ }d{{m}^{3}}$                 D)  $8\text{ }d{{m}^{3}}$

Given $\frac{{{P}_{2}}}{{{P}_{1}}}=2,\frac{{{T}_{2}}}{{{T}_{1}}}=2,\,{{V}_{1}}=4\,d{{m}^{3}},\,{{V}_{2}}=?$ From gas equation,                 $\frac{{{P}_{1}}\,{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}\,{{V}_{2}}}{{{T}_{2}}}$ or            $\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{P}_{2}}}{{{P}_{1}}}\times \frac{{{T}_{1}}}{{{T}_{2}}}$ $\therefore$  $\frac{4}{{{V}_{2}}}=2\times \frac{1}{2}=1$ $\therefore$  ${{V}_{2}}=4\,\,d{{m}^{3}}$