Question

A diatomic ideal gas is compressed adiabatically to $\dfrac {1}{32}$ of its initial volume. If the initial temperature of the gas is ${T}_{i}$ (in Kelvin) and the final temperature is $a{T}_{i}$, the value of a is
A. 4
B. 5
C. 2
D. 3

Answer 1

To solve this problem, use the adiabatic formula. Substitute the values for before compression and after compression in the adiabatic formula. Then, equate them. Substitute the value for atomicity of diatomic gas. It is given that the volume of the gas after compression is $\dfrac {1}{32}$ of its initial volume. So, substitute this condition in the above expression. Relation between the initial temperature and the final temperature is given so compare this given relation with the obtained equation. This will give the value of a.

Formula used:
$V{T}^ {\gamma -1}= constant$

Complete answer:
Let ${V}_{1}$ be the initial volume of the gas
${V}_{2}$ be the volume of the gas after compression
${T}_{f}$ be the final temperature of the gas
Given: ${V}_{2}= \dfrac {1}{32}{V}_{1}$
${T}_{f}= a{T}_{i}$
Adiabatic formula is given by,
$V{T}^ {\gamma -1}= constant$
So, we can write the above equation as,
${V}_{1}{T}_{1}^{\gamma -1} = {V}_{2}{T}_{2}^{\gamma -1}$
$\Rightarrow{V}_{1}{T}_{i}^{\gamma -1} = {V}_{2}{T}_{f}^{\gamma -1}$ ...(1)
We know, for a diatomic ideal gas, atomicity is given by,
$\gamma = \dfrac {7}{5}$
Substituting this value in the equation. (1) we get,
${V}_{1}{T}_{i}^{\dfrac {7}{5} -1} = {V}_{2}{T}_{f}^{\dfrac {7}{5} -1}$
$\Rightarrow {V}_{1}{T}_{i}^{\dfrac {2}{5}} = {V}_{2}{T}_{f}^{\dfrac {2}{5}}$
$\Rightarrow {V}_{1}{T}_{i}^{0.4}= {V}_{2}{T}_{f}^{0.4}$
Now, substituting value of ${V}_{2}$ in above equation we get,
${V}_{1}{T}_{i}^{0.4}= \dfrac {{V}_{1}}{32}{T}_{f}^{0.4}$
$\Rightarrow {T}_{i}^{0.4}= \dfrac {1}{32}{T}_{f}^{0.4}$
$\Rightarrow 32 {T}_{i}^{0.4}= {T}_{f}^{0.4}$ ...(2)
$\Rightarrow 4 {T}_{i}={T}_{f}$ ...(2)
But we know, ${T}_{f}= a{T}_{i}$ ...(3)
Comparing equation. (2) and (3) we get,
$a= 4$
Hence, the value of a is 4.

So, the correct answer is “Option A”.

Note:
To solve these types of problems, students should remember the atomicity of at least monoatomic, diatomic and triatomic gases. For monoatomic gases, the value of atomicity is $\dfrac{3}{2}$, while for triatomic gases, it is $\dfrac {6}{2}$. The numbers in the numerator of the value of atomicity describes the number of degrees of freedom. For example, monoatomic gases have 3 degrees of freedom.