# Question a4af0

May 28, 2015

The new volume of the gas will be equal to 560 mL.

You need to calculate the new volume of the gas knowing that both temperature and pressure change, which implies that you'll have to use the combined gas law equation

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

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

It's important to remember that the temperature of the gas must be expressed in Kelvin, not in degrees Celsius. Plug your values into the above equation and solve for ${V}_{2}$

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

V_2 = (3.0cancel("atm"))/(1.0cancel("atm")) * ((273.15 + 30.0)cancel("K"))/((273.15 + 50.0)cancel("K")) * "200.0 mL"#

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

Rounded to two sig figs, the number of sig figs you gave for the two pressures, the answer will be

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

Notice that the decrease in pressure offsets the decrease in temperature, and ultimately causes the final volume to be bigger than the initial volume.