# Question 21f80

Apr 30, 2015

The new volume of the sample will be 560 mL.

Because the number of moles is kept constant, you'll have to use the combined gas law to solve for the sample's new volume.

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

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and temperature at an initial state;
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and temperature at a final state.

Rearrange the above equation and solve for ${V}_{2}$ to get

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

V_2 = (3.0cancel("atm"))/(1.0cancel("atm")) * ((273.15 + 30)cancel("K"))/((273.15 + 50)cancel("K")) * "200.0 mL"#

${V}_{2} = \text{562.9 mL}$

Rounded to two sig figs, the answer will be

${V}_{2} = \textcolor{g r e e n}{\text{560 mL}}$