Question

According to Charles law:
(a) ${{(\dfrac{dV}{dT})}_{p}}=k$
(b) ${{(\dfrac{dV}{dT})}_{p}}=-k$
(c) ${{(\dfrac{dV}{dT})}_{p}}=-\dfrac{k}{T}$
(d) None of these

Answer 1

Hint: Charles Law states that pressure remaining constant, the volume of a given gas increases or decreases by 1/273 of its volume at ${{0}^{\circ }}C$ for every one-degree centigrade rise or fall in temperature. For solving numerical, the equation $\dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}}$ at constant pressure is used.

Complete step by step answer:
According to Charles Law:
 Pressure remaining constant, the volume of a given gas increases or decreases by 1/273 of its volume at ${{0}^{\circ }}C$ for every one-degree centigrade rise or fall in temperature.

Mathematically,
${{V}_{t}}={{V}_{0}}\left( \dfrac{273+t}{273} \right)$
Where ${{V}_{t}}$ is the volume at temperature t, and ${{V}_{0}}$ is the volume at ${{0}^{\circ }}C$
According to the Kelvin scale, T Kelvin is equal to Centigrade temperature + 273
$T\text{ K = }{{t}^{\circ }}C+273$
So, putting this value in the equation of the Charles Law, we get
${{V}_{t}}={{V}_{0}}\left( \dfrac{T}{273} \right)$

As we know that ${{V}_{0}}$ and 273 are constant, hence
${{V}_{t}}\propto T$, or simply we can say
$V\propto T$
So, we can say
$V=kT$

The numerical value of the constant k depends upon the number of gas taken and the pressure.
The above relation gives another definition of Charles Law as follows:
Pressure remaining constant, the volume of a given mass of a gas is directly proportional to its temperature in degree Kelvin.

The relation $V\propto T$ implies that,
$\dfrac{V}{T}=k$
Now, on differentiating the above equation at constant pressure, we get
${{\left( \dfrac{dV}{dT} \right)}_{p}}=k$
So, the correct answer is “Option A”.

Note: With the help of this relation, if ${{V}_{1}}$ is the volume of the gas at ${{T}_{1}}$ temperature and keeping the pressure constant, the temperature is changed to ${{T}_{2}}$, then the volume change to ${{V}_{2}}$ such that
$\dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}}$ at constant pressure.