Question

A constant pressure air thermometer gave a reading of 47.5 units of volume when immersed in ice cold water, and 67 units in a boiling liquid. The boiling point of the liquid will be
A. ${135^ \circ }C$
B. ${125^ \circ }C$
C. ${112^ \circ }C$
D. ${100^ \circ }C$

Answer 1

Hint: Since the pressure is given as constant, we can use Charles’s law to find the relationship between volume and temperature.
According to Charles’s law the volume is directly proportional to the temperature. By using this we can find the temperature of the boiling point of the liquid.

Complete answer:
It was given that at constant pressure an air thermometer gave a reading of 47.5 units when immersed in cold water.
Let this temperature of ice-cold water with initial temperature be denoted as ${T_1}$.
We know the temperature of ice is ${0^ \circ }$.
Thus,
$ \Rightarrow $${T_1} = 0 + 273 = 273\,K$
Let the volume at this temperature be denoted as ${V_1}$.
$ \Rightarrow $${V_1} = 47 \cdot 5\,\,$
When it is placed in a boiling liquid the reading is 67 units.
Let this volume be denoted as ${V_2}$.
$ \Rightarrow $ ${V_2} = 67$
we need to find the temperature of the boiling liquid.
At constant pressure the relationship between volume and temperature is given by the Charles law.
According to Charles law volume is directly proportional to temperature at constant pressure.
$ \Rightarrow $$V \propto T$
$ \Rightarrow V = kT$
Where V is the volume and T is the temperature and k is a constant.
So, for the initial case when the thermometer is placed in cold water, we can write
$ \Rightarrow $${V_1} = k{T_1}$ ……………………...(1)
For the second case when the thermometer is placed in boiling water, we can write
$ \Rightarrow $${V_2} = k{T_2}$ ……………………...(2)
By dividing equation 1 with 2, we get
$ \Rightarrow $$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{k{T_1}}}{{k{T_2}}}$
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$
$ \Rightarrow {T_2} = \dfrac{{{V_2}}}{{{V_1}}} \times {T_1}$
On substituting the values, we get
$ \Rightarrow {T_2} = \dfrac{{67}}{{42 \cdot 5}} \times 273$
$\therefore {T_2} = 385\,K$
This is the temperature in kelvin. To get the value in degree Celsius we need to subtract 273.
Thus,
$ \Rightarrow $${T_2} = 385\, - 273 = {112^ \circ }C$
This is the temperature of boiling point of the liquid.

So, the correct answer is option C.

Note:
Charles's law which states that volume is directly proportional to temperature is only applicable at constant pressure. The proportionality constant k that appears in the equation depends on the amount of substance and the constant pressure applied. So, if pressure is changing, we cannot apply Charles’s law.