# Question 5609e

Jan 2, 2015

I'm assuming that 616 mg is actually 616 mmHg, the gas' pressure.

You can approach this problem by using the combined gas law, since the number of moles does not change.

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

You know that

${V}_{1} = 322$ "mL,
${T}_{1} = {29}^{\circ} C = 273.15 + 29 = 302.15$ $\text{K}$
${P}_{1} = 616$ $\text{mmHg} = 616$ "mmHg" * ("1 atm")/(760"mmHg") = 0.810# $\text{atm}$

At STP, the temperature is equal to ${T}_{2} = 273.15$ $\text{K}$ and pressure is equal to ${P}_{2} = 1.00$ $\text{atm}$. Plug all in and get

${V}_{2} = {P}_{1} / \left({P}_{2}\right) \cdot {T}_{2} / {T}_{1} \cdot {V}_{1} = \frac{0.810 a t m}{1.00 a t m} \cdot \frac{273.15 K}{302.15 K} \cdot 322 m L$

${V}_{2} = 236$ $\text{mL}$