Question

The work of 146 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 ${{7}^{\circ }}C$. The gas is ($R=8.3Jmo{{l}^{-1}}{{K}^{-1}}$)
(A) Diatomic
(B) Triatomic
(C) A mixture of monoatomic and diatomic
(D) Monoatomic

Answer 1

In adiabatic compression, the temperature rises and in the case of expansion which is against pressure, causes a drop in the temperature. Compression leads to decrease in volume that causes an increase in pressure and this leads to increase in temperature. This is due to the conversion of work energy to internal energy.

Formulas used:
Work done in adiabatic process $W=\dfrac{nR({{T}_{2}}-{{T}_{1}})}{\gamma -1}$
Where $n$ is the amount of substance, $R$ is the ideal gas constant, $({{T}_{2}}-{{T}_{1}})$ is the difference in temperature, and $\gamma$ is the ratio between the molar heat capacities.

Complete step by step answer:
We are given that the increase in temperature $({{T}_{2}}-{{T}_{1}})$ is ${{7}^{\circ }}C$
The work done on the system, $W=146kJ=146\times 1000J$
The number of moles of gas, $n=1000$
We know that the work done in an adiabatic process is
$W=\dfrac{nR({{T}_{2}}-{{T}_{1}})}{\gamma -1}$
Upon substituting the given values, we obtain the following:
$146000=\dfrac{1000\times 8.3\times 7}{\gamma -1}$
$\begin{aligned} & \Rightarrow \gamma -1=0.39 \\\ & \therefore\gamma \approx 1.4 \\\ & \\\ \end{aligned}$
$\gamma$values are capable of determining the speed of sound in a gas during adiabatic processes. When its value is 1.66, we consider the gas to be ideally monoatomic and when it is 1.4, we can say that it is diatomic.
Hence, the given gas is diatomic.
Therefore, Option A is the right answer among the four options.

Note: The particles that form an ideal gas are considered to be point-like elements without a detailed structure. So, the real gases that are similar to this particular ideal model should mostly be monatomic, with the smallest of atoms.