Question

Two flasks A and B have equal volumes. A is maintained at $300K$ and B at $600K$ , while A contains ${H_2}$ gas, B has an equal mass of $C{O_2}$ gas. Find the ratio of total K.E of gases in flask A to that of B.
A.$1:2$
B.$11:1$
C.$33:2$
D.$55:7$

Answer 1

To answer this question, you should know about the concept of kinetic theory of gases. This theory tells about the thermodynamic behavior of gases. Here, it is given that the mass of gases is equal. We have to just put the formula of kinetic energy of gases and find its ratio.

Complete answer:
Kinetic theory of gases tells about the thermodynamic behavior of gases. It explains the macroscopic properties of gases such as volume, pressure and temperature. It also tells about the transport properties such as viscosity, thermal conductivity and mass diffusivity.
The kinetic theory of gases make some basic assumptions such as:
A) The molecules of the gases do not interact with each other.
B) The elastic collision occurs in molecules with themselves and with the wall of the container.
C) The momentum and kinetic energy is conserved.
The formula for kinetic energy of gases is as follows:
$K.E.\,\, = \,\dfrac{{3\,nRT}}{2}$
Here, given mass of both gases i.e. ${H_2}$ and $C{O_2}$ is same which is assume to be ‘m’
We know that,
$n\, = \dfrac{{Given\,\,mass}}{{Molar\,\,mass}}$
For flask A i.e. ${H_2}$gas
Temperature $({T_A})$ = $300K$
${n_A}\, = \dfrac{m}{2}$
So, $K.E.{\,_A}\,\, = \,\dfrac{{3\,{n_A}\,R\,{T_A}}}{2}$
Put the value of ${n_A}$ and ${T_A}$
$K.E.{\,_A}\,\, = \,\dfrac{{3\,m\,R\,(300)}}{4}$ …. (a)
Similarly, for flask B i.e. $C{O_2}$ gas
Temperature $({T_B})$ = $600K$
${n_B}\, = \dfrac{m}{{44}}$
So, $K.E.{\,_B}\,\, = \,\dfrac{{3\,{n_B}\,R\,{T_B}}}{2}$
Put the value of ${n_B}$ and ${T_B}$
We get,
$K.E.{\,_B}\,\, = \,\dfrac{{3\,m\,R\,(600)}}{{88}}$ …. (b)
Divide (a) by (b)
$\dfrac{{K.E.{\,_A}}}{{K.E.{\,_B}}}\,\, = \dfrac{{3\,m\,R\,(300)}}{4}\,\, \times \,\,\dfrac{{88}}{{3\,m\,R\,(600)}}\,$
After solving the above equation we get,
$\dfrac{{K.E.{\,_A}}}{{K.E.{\,_B}}}\,\, = \dfrac{{300}}{4}\,\, \times \,\,\dfrac{{88}}{{600}}\,$
$\dfrac{{K.E.{\,_A}}}{{K.E.{\,_B}}}\,\, = \dfrac{{11}}{1}\,\,$
Therefore, the ratio of total K.E of gases in flask A to that of B is $11:1$

Hence, the correct answer is option (B).

Note:
Remember, the significance of kinetic theory of gases is that by knowing the temperature, we can find out the average kinetic energy of a gas molecule. It does not matter what gas we are considering unless and until it is an ideal gas. We can also calculate the microscopic parameters like momentum, velocity, internal energy, thermal energy etc.