Courses
Courses for Kids
Free study material
Offline Centres
More

# The temperature of the gas contained in a closed vessel increases by ${1^0}C$ when pressure of the gas is increased by $1\%$, the initial temperature of the gas is:A. $100\,K$B. ${100^0}C$C. $200\,K$D. ${250^0}C$

Last updated date: 22nd Feb 2024
Total views: 304.2k
Views today: 9.04k
Verified
304.2k+ views
Hint:In theory of gases and molecules, whenever the pressure of a gas increased then the temperature of the gas also varies linearly with the pressure and this law in mathematical formulation is written as $P = kT$ where $k$ is the proportionality constant.

Complete step by step answer:
Let us imagine the initial pressure of the gas is $P$ an increase in pressure is $1\%$ which can be represented as,
$P' = P + \dfrac{P}{{100}}$
Where $P'$ is the new pressure of the gas.
$\Rightarrow P' = 1.01P$
$\Rightarrow \dfrac{{P'}}{P} = 1.01$
Now, Let us assume that the initial temperature of the gas when pressure was $P$ is $T$ and it’s given that the temperature is increased by ${1^0}C$ which means new temperature at new pressure $P'$ is $T'$ and it’s given by,
$T' = T + 1$

Now using the law of $P = kT$ we can compare this law with initial and final values of temperature and pressure as
$\dfrac{{P'}}{P} = \dfrac{{T'}}{T}$
Putting the values we will get
$1.01 = \dfrac{{T + 1}}{T}$
$\Rightarrow 0.01\,T = 1$
$\Rightarrow T = 100K$
So, the initial temperature of the gas when its pressure gets increased by one percentage is $\therefore T = 100K$

Hence, the correct option is A.

Note: It should be remembered that, the law governing that pressure is directly proportional to the temperature of the gas is subject to when volume of the gas remains constant and this particular law in theory of gases is known as Gay Lussac’s Law which is $P \propto T$.