A 780*mL volume of gas 1.00*atm pressure, and 310*K temperature, is cooled to 295*K, and 65*mm*Hg pressure. What is the new volume?

Nov 14, 2016

For a given quantity of gas, .........${V}_{2} = 8.70 \cdot L .$

Explanation:

For a given quantity of gas, the combined gas law holds that $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, temperature quoted in $\text{degrees Kelvin}$.

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

=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)

$= 8.69 \times {10}^{3} \cdot m L = 8.70 \cdot L$

We use here the useful relationship that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$.

And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.

Please note that I used $22$ ""^@C, $\equiv$ $295 \cdot K$, and $37$ ""^@C, $\equiv$ $310 \cdot K$. You can change the calculation accordingly if I have got your starting conditions wrong.