Question

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!

Answer 1

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:

$\dfrac{nRT}{nRT}=\dfrac{808\times10^3}{(213\times10^3)\cdot V_2}\\\\ 1=\dfrac{808}{213\cdot V_2}\\\\ 213V_2=808\\\\ V_2=\dfrac{808}{213}\\\\ \boxed{V_2\approx 3.79~L}$

Answer 2

Hello! 

Data:

Vo (initial volume) = 8 L
V (final volume) = ?
Po (initial pressure) = 101 kPa
P (final pressure) = 213 kPa 

We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the formula, we have:

$P_0*V_0 = P*V$

$101*8 = 213*V$

$808 = 213V$

$213V = 808$

$V = \dfrac{808}{213}$

$\boxed{\boxed{V \approx 3,79\:L}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR!