# Question d9f89

Apr 29, 2015

The new volume wil be 3 L.

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

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

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

Make sure that you use the same units for pressure and that you convert the temperature from degrees Celsius to Kelvin

V_2 = (1.00cancel("atm"))/(150/760cancel("atm")) * ((273.15)cancel("K"))/((273.15 + 36)cancel("K")) * "0.600 L"#

${V}_{2} = \text{2.685 L}$

Rounded to one sig fig, the number of sig figs given for ${0}^{\circ} \text{C}$, the answer will be

${V}_{2} = \textcolor{g r e e n}{\text{3 L}}$