Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

One mole of a perfect gas in a cylinder fitted with a piston has a pressure $P$ , volume $V$ and temperature $T$ . The temperature is increased by $1K$ keeping pressure constant, the increase in volume is?
A. $\dfrac{{2V}}{{273}}$
B. $\dfrac{V}{{91}}$
C. $\dfrac{V}{{273}}$
D. $V$

Answer
VerifiedVerified
163.2k+ views
Hint: It is given that, the temperature is increased by 1K at constant pressure. This implies that, the ratio of volume to temperature i.e., \[\dfrac {V}{T}\] will be same before and after increasing the temperature. So we can compare the ratio as \[\dfrac {V_1}{T_1} = \dfrac {V_2}{T_2}\]. Substitute all the given values in the equation for finding \[V_2\].

Complete answer:
It is given that $1$ mole of gas is fitted with a piston inside a cylinder with:
An initial pressure of ${P_1} = P$ , initial volume of ${V_1} = V$ , and initial temperature of ${T_1} = T$
Assume when the temperature is increased by $1K$ at constant pressure, ${T_2} = T + 1$

But, if we consider NTP thermodynamic conditions, the absolute temperature is $T = 273K$ i.e.,
${T_1} = T = 273K$ and ${T_2} = T + 1 = 273 + 1 = 274K$

Now, we know that at constant pressure, the volume of gas is directly proportional to its absolute temperature (according to Charles Law).
i.e., $\dfrac{V}{T} = $constant
or, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ … (1)

Substitute the values of ${T_1}$ , ${T_2}$ and ${V_1}$ given in the question in eq. (1), we get
$ \Rightarrow \dfrac{V}{{{V_2}}} = \dfrac{{273}}{{274}}$
$ \Rightarrow {V_2} = \dfrac{{274}}{{273}}V$

And, increase in volume will be: -
$Increase = {V_2} - {V_1} = \dfrac{{274}}{{273}}V - V$
i.e., $Increase = \dfrac{V}{{273}}$

Thus, when the temperature is increased by $1K$ at constant pressure, the increase in volume is $\dfrac{V}{{273}}$.

Hence, the correct option is (C) $\dfrac{V}{{273}}$ .

Note: It is important to note that the initial value of the temperature must be considered as 273K. When the initial value of the temperature is not provided. We should consider the initial value of the temperature according to NTP standard.