Question

A vertical hollow cylinder of height $152$cm is fitted with a weightless piston. The lower half is filled with an ideal gas and the upper half with mercury. The cylinder is now heated at $300K$, so that $\dfrac{1}{4}$th of mercury comes out. Find the temperature at which it will happen assuming thermal expansion of Hg to be negligible
A. $340K$
B. $330.6K$
C. $328.12K$
D. $322.2K$

Answer 1

Hint: First find the pressure and volume at initial stage( before heating) using Boyle's law which states that volume is inversely proportional to the pressure of the gas at constant temperature. Then find the pressure and volume at the final stage. After that use gas equation for ideal gas which is given as-
$\dfrac{{PV}}{{RT}} = 1$ to find the temperature at the final stage.

Step-by-Step Solution-
Given, a vertical hollow cylinder fitted with a weightless piston has height= $152$cm.
The lower half is filled with ideal gas and the upper half with mercury.
We know that at atmospheric pressure the volume of mercury is $76$cm and the standard pressure of atmospheric air is $76$cm.
Then at initial stage at lower half
 Pressure of ideal gas ${{\text{P}}_{\text{1}}}$=Pressure of Hg P+ Pressure of atmospheric air${{\text{P}}_0}$
On putting values we get,
Pressure of ideal gas=$76 + 76 = 152$ cm
So, Pressure of the ideal gas is twice the pressure of mercury.
$ \Rightarrow {{\text{P}}_{\text{1}}} = 2{\text{P}}$ -- (i)
Then we know from Boyle’s law that volume is inversely proportional to the pressure of the gas at constant temperature.
Then at temperature ${T_1}$ of $300K$,
$V \propto \dfrac{1}{P}$ -- (ii)
Then from (i) and (ii) we can write,
\[ \Rightarrow \dfrac{1}{{{V_1}}} = \dfrac{2}{V}\]
On solving the above we get,
$ \Rightarrow {V_1} = \dfrac{V}{2}$
This is the volume of the ideal gas.
Now after heating $\dfrac{1}{4}$th of mercury comes out so the pressure of mercury changes.
Then pressure of mercury= $57$ cm
Then at final stage,
The Pressure of the ideal gas
Pressure of ideal gas P=Pressure of Hg + Pressure of atmospheric air ${{\text{P}}_0}$
Pressure of ideal gas P= $57 + 76 = 133$ cm
So according to Boyle’s law the volume of the ideal gas will become five eighth of the initial volume of mercury
$ \Rightarrow {V_2} = \dfrac{{5V}}{8}$
We have to find the temperature $T$ at final stage.
From gas equation we know that $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{T}$ for ideal gas then putting the values of Pressure, Volume and temperature of initial and final stage on the given formula and equating both we get,
$ \Rightarrow \dfrac{{152 \times {V_1}}}{{2 \times 300}} = \dfrac{{133 \times 3{V_1}}}{{4 \times T}}$
On solving we get,
$ \Rightarrow T = \dfrac{{133 \times 300 \times 3}}{{152 \times 2}}$
On simplifying we get,
$ \Rightarrow T = 328.12$K

Answer-The correct answer is C.

Note: Generally the formula is $\dfrac{{PV}}{{nRT}} = Z$ for real gases, where n is the number of moles and R is gas constant. The value of Z is decided on following points-
- Z is equal to one for ideal gases while for all other gases it is less than one at low pressure.
- Z is greater than one for all the other gases at high pressure and hydrogen and helium.