Question

When the temperature of 23 mL $C{{O}_{2}}$ gas is changed from $10{}^\circ $ to $30{}^\circ $ at constant pressure of 760 mm then what will be the volume of gas.

Answer 1

Hint: the properties of gases can be studied by various gas laws. The laws include Boyle’s law, Charles’ law, Gay Lussac’s law, Avogadro’s law, etc. these laws have various relationships between pressure, temperature, volume and number of moles of a gas. At constant pressure, the volume and temperature relationship is given by Charles’ law.

Complete answer:
We have been given a gas carbon dioxide$C{{O}_{2}}$, with volume 23 mL. At constant pressure of 760 mm, the temperature of this gas changes from$10{}^\circ $ to $30{}^\circ $ . We have to find the volume at the final temperature change.
The properties of gases are studied through various gas laws. The gas law that includes the temperature and volume change at constant pressure is Charles' law. According to this law, at constant pressure the volume of a certain mass of a gas is directly proportional to the temperature, that is
$V\propto T$ , V is volume and T is temperature.
Through this law we can derive an expression that $V = kT$, where $k$ is a constant of proportionality.
So, $k=\dfrac{V}{T}$
Considering two gases at temperature, ${{T}_{1}}$ and ${{T}_{2}}$ having volumes${{V}_{1}}$ and ${{V}_{2}}$, we have the relationship$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$ , putting the given values in this formula we have,
$\dfrac{23}{{{V}_{2}}}=\dfrac{10+273}{30+273}$
$\dfrac{23}{{{V}_{2}}}=\dfrac{283}{303}$
${{V}_{2}}=\dfrac{303\times 23}{283}$
${{V}_{2}}=24.6\,mL$
Hence, the volume of $C{{O}_{2}}$ will be 24.6 mL.

Note:
The calculations involving temperature in gas laws always contain the temperature in Kelvin, therefore the given Celsius temperature is changed in Kelvin by the conversion factor $0{}^\circ C=273\,K$. The application of Charles’ law is used in the hot air balloons, as hot air is less dense so the balloon rises up.