Question

The work of $146\,{\text{kJ}}$ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by $7^\circ {\text{C}}$. The gas is $\left( {R = 8.3\,{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}} \right)$ :
(A) Diatomic
(B) Triatomic
(C) A mixture of monoatomic and diatomic
(D) Monoatomic

Answer 1

First of all, we will find an expression for energy in an adiabatic process which also contains a ratio of specific heats. We will substitute the required values and manipulate accordingly.

Complete step by step answer:
In the given, question we are supplied with the following data:
The total work done is $146\,{\text{kJ}}$ .
Number of moles $\left( n \right)$ is one kilo moles.
The increase in temperature is $7^\circ {\text{C}}$ .
Gas constant is given as $R = 8.3\,{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
We know,
$1\,{\text{kJ}} = 1000\,{\text{J}}$
So,
$146\,{\text{kJ}} \\\ = {\text{146}} \times {\text{1}}{{\text{0}}^{\text{3}}}\,{\text{J}} \\\ = 146000\,{\text{J}} \\\$
The temperature change can also be written as:
$\Delta T = 7^\circ {\text{C}}$
We are asked to find out what is the nature of the gas. For this we need to find the ratio of the specific heats which is represented by a Greek symbol $\left( \gamma \right)$. If the ratio of the specific heats comes out to be $1.6$, then the gas is ideal monatomic and if the ratio of the specific heats comes out to be $1.4$, then the gas is a diatomic one.
We have a formula which gives the work done in adiabatic process, as follows:
$W = \dfrac{{nR\left( {{T_2} - {T_1}} \right)}}{{\gamma - 1}}$ …… (1)
Where,
$W$ indicates the work done on the gas.
$n$ indicates the number of moles.
$R$ indicates gas constant.
${T_2}$ indicates final temperature.
${T_1}$ indicates initial temperature.
$\gamma$ indicates ratio of specific heats.
Now, we substitute the required values in the equation (1) and we get:
W = \dfrac{{nR\Delta T}}{{\gamma - 1}} \\\
146000 = \dfrac{{1000 \times 8.3 \times 7}}{{\gamma - 1}} \\\
\gamma - 1 = \dfrac{{1000 \times 8.3 \times 7}}{{146000}} \\\
\gamma - 1 = 0.398 \\\
We now further simplify:
\Rightarrow \gamma - 1 = 0.398 \\\
\Rightarrow \gamma = 1 + 0.398 \\\
\Rightarrow \gamma = 1.398 \\\
\Rightarrow \gamma \sim 1.4 \\\
Since, the value of ratio of specific heats comes out to be $1.4$, so the gas is a diatomic.
The correct option is (A).

Note: In the given problem, we need to take care of the units. It always needs to be in S.I units. To correctly answer this question, we need to remember the value of the ratio of the specific heats for different types of gases whether it be monatomic or diatomic. In an adiabatic system there is no transfer of heat or mass between the system and the surrounding.