Question

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be \[n\] times the initial pressure. The value of \[n\] is:
A. 32
B. \[\dfrac{1}{32}\]
C. 326
D. 128

Answer 1

Hint: We will use the concept of adiabatic process to solve the question which states that an adiabatic process is the thermodynamic process in which neither during expansion nor compression does heat from the system transfer to its surroundings. Both reversibility and irreversibility are options for the adiabatic process.

Formula used:
The adiabatic process equation is as follows:
\[P{V^y} = {\rm{ constant}}\],
P is the system's pressure.
V is the system’s volume.
And y is the ratio of the heat capacity at constant volume C_v to the heat capacity at constant pressure C_p is known as the adiabatic index.

Complete Step by Step Solution:
In adiabatic process
\[P{V^\gamma } = {\rm{ constant}}\]
\[\therefore P{\left( {\dfrac{m}{\rho }} \right)^\gamma } = \] constant
As mass is constant
\[\therefore P \propto {\rho ^\gamma }\]
If \[{P_i}\] and \[{P_f}\] be the initial and final pressure of the gas and \[{\rho _i}\] and \[{\rho _f}\] be the initial and final density of the gas. Then
\[\dfrac{{{P_f}}}{{{P_i}}} = {\left( {\dfrac{{{\rho _f}}}{{{\rho _i}}}} \right)^\gamma } = {(32)^{7/5}}\]
\[ \Rightarrow \dfrac{{n{P_i}}}{{{P_i}}} = {\left( {{2^5}} \right)^{7/5}} = {2^7}\]
\[ \Rightarrow n = {2^7} = 128\].

Hence, option D is the correct answer.

Note: Students can make mistakes in substituting the formula of V which will result in getting the wrong answer. The prefix -di in diatomic gases signifies a combination or a value of two, and atomic indicates that the prefix is to be applied for atoms \[({O_2},{N_2},{H_2},etc)\]. The chemical properties of the elements that make up those combinations lead to the state of matter in which those combinations are found.