# A container of oxygen has a volume of 349 mL at a temperature of 22.0°C . What volume will the gas occupy at 50.0°C?

Nov 26, 2015

We assume a constant pressure. Thus Charles' Law applies.

#### Explanation:

Charles' Law states that at constant pressure, and constant amount of gas, volume is proportional to absolute temperature, i.e. $V \propto T$.

So $V = k T$, where $V$ = volume, $k$ is some constant, and $T$ is absolute temperature.

So, in each case, ${V}_{1} = k {T}_{1} , \mathmr{and} {V}_{2} = k {T}_{2}$. If we solve for $k$, then ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$.

Thus, ${V}_{2} = \frac{{V}_{1} \times {T}_{2}}{T} _ 1$ $=$ $\frac{0.349 \cdot {\mathrm{dm}}^{3} \times 337 \cancel{K}}{295 \cancel{K}}$ $=$ ?? ${\mathrm{dm}}^{3}$

Note that ${V}_{2}$ will be bigger (reasonably), because ${T}_{2}$ has increased from ${T}_{1}$.

Nov 26, 2015

The final volume will be 382 mL.

#### Explanation:

Charles' law states that the volume of a given amount of a gas kept at constant pressure, is directly proportional to the temperature in Kelvins. http://chemistry.bd.psu.edu/jircitano/gases.html

The equation needed to solve this problem is ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$.

Given/Known
${V}_{1} = \text{349 mL}$
${T}_{1} = \text{22.0"^"o""C"+273.15="295.2 K}$
${T}_{2} = \text{50.0"^"o""C"+273.15="323.2 K}$

Unknown
${V}_{2}$

Equation

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Solution
Rearrange the equation to isolate ${V}_{2}$ and solve.

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

V_2=(349"mL"xx323.2cancel"K")/(295.2cancel"K")="382 mL"