
$300\,ml$ of a gas at ${{27}^{{\mathrm O}}}C$ is cooled to $-{{3}^{{\mathrm O}}}C$ at constant pressure, the final volume is
A.$540\,ml$
B.$135\,ml$
C.$270\,ml$
D.$350\,ml$
Answer
163.5k+ views
Hint: At constant pressure, the volume occupied by the gas molecules is directly proportional to the temperature. In the given problems the temperature of the system decreases at constant pressure therefore according to Charles’ law volume of the gas will also be changed.
Formula Used: We can write the mathematical expression of Charles’ law in the following way,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Here ${{V}_{1}}=$Initial volume of gas molecules
${{V}_{2}}=$Final volume of gas molecules
${{T}_{1}}=$Initial temperature
${{T}_{2}}=$Final temperature
Complete step by step solution:Charles’ law provides a direct relationship between the volume of gas and the temperature of that system. According to this law, the temperature,$T$ of the system is directly proportional to the volume,$V$ occupied by the gas molecules at constant pressure.
$\therefore \,\,\,\,\,\,V\,\,\alpha \,\,T$
Charles’ law is useful to anticipate the effect of changes on temperature and volume. If under reaction conditions initial volume,${{V}_{1}}$and initial temperature,${{T}_{1}}$change to the final volume,${{V}_{2}}$and final temperature,${{T}_{2}}$. Then we can apply Charles’ law here and it can be expressed in the following way:
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Or,${{V}_{2}}={{V}_{1}}\,\times \dfrac{{{T}_{1}}}{{{T}_{2}}}$ ……(i)
Given, ${{V}_{1}}=300\,\,ml$, ${{T}_{1}}=(273+27)={{300}^{o}}K$
And ${{T}_{2}}=(273-3)={{270}^{o}}K$
Putting these values in equation (i),
${{V}_{2}}=300\,\,ml\,\times \dfrac{{{270}^{o}}K}{{{300}^{o}}K}$
Or, ${{V}_{2}}=270\,\,ml$
Therefore the final volume of the gas molecules is $270\,\,ml$ .
Thus option (C) is correct.
Note: Though perfect gas follows Boyle’s law, Charles’ Law, and Avogadros’ law at low pressure and high temperature, but deviates at high pressure and low temperature. The reason behind this fact is that at high-pressure gas molecules are closer together as a result intermolecular forces are stronger and at lower temperatures, gas molecules would not possess any random motion due to lower volumes of gasses.
Formula Used: We can write the mathematical expression of Charles’ law in the following way,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Here ${{V}_{1}}=$Initial volume of gas molecules
${{V}_{2}}=$Final volume of gas molecules
${{T}_{1}}=$Initial temperature
${{T}_{2}}=$Final temperature
Complete step by step solution:Charles’ law provides a direct relationship between the volume of gas and the temperature of that system. According to this law, the temperature,$T$ of the system is directly proportional to the volume,$V$ occupied by the gas molecules at constant pressure.
$\therefore \,\,\,\,\,\,V\,\,\alpha \,\,T$
Charles’ law is useful to anticipate the effect of changes on temperature and volume. If under reaction conditions initial volume,${{V}_{1}}$and initial temperature,${{T}_{1}}$change to the final volume,${{V}_{2}}$and final temperature,${{T}_{2}}$. Then we can apply Charles’ law here and it can be expressed in the following way:
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Or,${{V}_{2}}={{V}_{1}}\,\times \dfrac{{{T}_{1}}}{{{T}_{2}}}$ ……(i)
Given, ${{V}_{1}}=300\,\,ml$, ${{T}_{1}}=(273+27)={{300}^{o}}K$
And ${{T}_{2}}=(273-3)={{270}^{o}}K$
Putting these values in equation (i),
${{V}_{2}}=300\,\,ml\,\times \dfrac{{{270}^{o}}K}{{{300}^{o}}K}$
Or, ${{V}_{2}}=270\,\,ml$
Therefore the final volume of the gas molecules is $270\,\,ml$ .
Thus option (C) is correct.
Note: Though perfect gas follows Boyle’s law, Charles’ Law, and Avogadros’ law at low pressure and high temperature, but deviates at high pressure and low temperature. The reason behind this fact is that at high-pressure gas molecules are closer together as a result intermolecular forces are stronger and at lower temperatures, gas molecules would not possess any random motion due to lower volumes of gasses.
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