# If the pressure of 50.0 mL of oxygen gas at 100°C increases from 735 mm Hg to 925 mm Hg, what is the final volume? Assume the temperature remains constant.

Feb 16, 2017

$\text{volume} = 0.03976 L$

#### Explanation:

Recall the gas law

$P V = n R T$

$P = \frac{n R T}{V}$

Our question
If the pressure of 50.0 mL of oxygen gas at 100°C increases from 735 mm Hg to 925 mm Hg, what is the final volume? Assume the temperature remains constant.
This means

735mmHg = "{n * 0.0821L * (100C + 273K )}"/ ("50.0mL "/"1000ml" )

mmHg must be converted in $a t m$ , $\text{celsius to kelvin}$ and $\text{ml to Litres}$

=1 atm = 760mmHg

$\text{735mmHg" /"760mmHg} = 0.96711 a t m$

0.9671053242406501atm = n * 0.0821L * 373K/0.05L

Solve the equation

$0.9671053242406501 a t m = \text{30.6233n"/"0.05L}$

Multiply both sides with 0.05L

$\setminus \frac{30.6233 n}{0.05 L} \cdot 0.05 = 0.96711 a t m \cdot 0.05 L$

$30.6233 n = 0.0483555$

$n = \frac{30.6233}{0.048355}$

n = 0.00158moles

So lets now calculate volume when we know n.
n is constant in both the equation.

$\text{925mmHg"/"760mmHg} = 1.2171052631578947368421052631579 a t m$

$1.21711 a t m = \frac{\text{(0.00158 * 0.0821L * 373K)}}{V}$

$1.21711 a t m = \frac{0.048384814}{V}$

$V = \frac{0.048384814}{\text{1.21711atm}}$

$V = 0.03976 L$

$\text{volume} = 0.03976 L$