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?

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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?

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Answer 1 · 2015-11-26T18:16:14

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. $V \propto T$ $VpropT$ .

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

So, in each case, $V_{1} = k T_{1}, and V_{2} = k T_{2}$ $V_1=kT_1, and V_2=kT_2$ . If we solve for $k$ $k$ , then $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1 = V_2/T_2$ .

Thus, $V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$ $V_2=(V_1xxT_2)/T_1$ $=$ $=$ $(0.349*dm^3xx337cancelK)/(295cancelK)$ $=$ $??$ $dm^3$

Note that $V_2$ will be bigger (reasonably), because $T_2$ has increased from $T_1$ .

Answer 2 · 2015-11-26T20:24:56

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
$V_1="349 mL"$
$T_1="22.0"^"o""C"+273.15="295.2 K"$
$T_2="50.0"^"o""C"+273.15="323.2 K"$

Unknown
$V_2$

Equation

$V_1/T_1=V_2/T_2$

Solution
Rearrange the equation to isolate $V_2$ and solve.

$V_2=(V_1T_2)/T_1$

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

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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?

2 Answers

Explanation:

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