Bowline347
37

# The gas in a cylinder has a volume of 8 liters at a pressure of 101 kPa. The pressure of the gas is increased to 213 kPa. Assuming the temperature remains constant, what would the new volume be? Please show your work, thanks!

Alex921

We'll use the Ideal Gas Law. Before: $p_1V_1=n_1RT_1\\\\ (101\times10^{3})\cdot8=n_1RT_1\\\\ n_1RT_1=808\times10^3~~~(i)$ After: $p_2V_2=n_2RT_2\\\\ (213\times10^3)\cdot V_2=n_2RT_2~~~(ii)$ Since the quantity of gas wasn't changed, $n_1=n_2=n$. Furthermore, the temperature remains constant, so $T_1=T_2=T$. Then, the expressions that were find are: $nRT=808\times10^3~~~(i)\\\\ nRT=(213\times10^3)\cdot V_2~~~(ii)$ Dividing them: $\dfrac{nRT}{nRT}=\dfrac{808\times10^3}{(213\times10^3)\cdot V_2}\\\\ 1=\dfrac{808}{213\cdot V_2}\\\\ 213V_2=808\\\\ V_2=\dfrac{808}{213}\\\\ \boxed{V_2\approx 3.79~L}$