Question

An open vessel is kept at 127°C. If this vessel is heated to
227°C then the fraction of air which escaped from the
vessel is
1) 3/4
2)1/5
3)2/3
4)1/4​

Answer 1

Answer:

• The fraction of the air escaped from the vessel = 1/5

OPTION (2) 1/5 is correct.

Explanation:

Given:-

Initial temperature, $\rm{T_1}$ = 127° C

New temperature, $\rm{T_2}$ = 227° C

To find:-

The Fraction of air which escaped on changing the temperature

Knowledge required:-

• Charle's law:  It states that the volume of a gas at constant pressure is directly proportional to the absolute temperature (in Kelvin). Such that, V/T = constant.

For a gas having initial values of Temperature and volume $\rm{T_1}$ and $\rm{V_1}$, and if the temperature is changed to $\rm{T_2}$ such that the volume becomes $\rm{V_2}$ then,

$\qquad \rm{\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}}$

Solution:-

Converting the temperature given in degree celsius into Kelvin.

$\rm{T_1 = 127^{\circ}\;C = 127 + 273 \;K = 400\;K}$

$\rm{T_2 = 227^{\circ}\;C = 227 + 273 \;K = 500\;K}$

Now Let, the initial volume of air in the vessel be $\rm{V_1}$ and the new volume be $\rm{V_2}$

then,

Initially, the fraction of air escaped from the vessel will be = $\rm{{\bigg(\dfrac{V_1}{V_2}\bigg)}_{initial}}$ = 1  

Using the Charle's law

calculating the new fraction of air escaped $\rm{{\bigg(\dfrac{V_1}{V_2}\bigg)}_{new}}$ =?

$\to \rm{\dfrac{V_1}{T_1}= \dfrac{V_2}{T_2}}$

$\to \rm{\dfrac{V_1}{V_2}= \dfrac{T_1}{T_2}}$

$\to \rm{\dfrac{V_1}{V_2}= \dfrac{400}{500}=\dfrac{4}{5}}$

Hence, $\rm{{\bigg(\dfrac{V_1}{V_2}\bigg)}_{new}}$ =  $\dfrac{4}{5}$

Therefore,

the fraction of air escaped = $\rm{{\bigg(\dfrac{V_1}{V_2}\bigg)}_{initial} -{\bigg(\dfrac{V_1}{V_2}\bigg)}_{new}}$ = $\rm{ 1 - \dfrac{4}{5} = \dfrac{1}{5}}$

 So,

• The Fraction of air escaped from the vessel is 1/5

Answer 2

Answer:

The Fraction of air escaped from the vessel is 1/5