Question

An ideal monoatomic gas is carried around the cycle ABCDA as shown in fig. The efficiency of the gas is:

A. $\dfrac{4}{21}$
B.$\dfrac{2}{21}$
C. $\dfrac{4}{31}$
D. $\dfrac{2}{31}$

Answer 1

Hint: The efficiency of the gas cycle is given by the ration of work done by the system to the heat absorbed from the source.

Complete step by step answer:
While looking at the figure we can see that the path from A to B is an Isochoric process where the volume process remains constant. So the work done is zero during this process since there is no change in volume. So the heat absorbed is given by the equation
$\text{Q}=\text{n}{{\text{C}}_{\text{v}}}\Delta T$
Where, ${{\text{C}}_{\text{V}}}$ is the specific heat of a monoatomic gas at constant volume, whose value is $\dfrac{3}{2}\text{R}$.
$\therefore \text{ Q}=\dfrac{3}{2}(\text{nR}\Delta T)$
We know that from the ideal gas equation, PV=nRT, so substituting this in above equation we get,
$\text{Q}=\dfrac{3}{2}\left( \text{V}\Delta P \right)$
$\Rightarrow \text{Q}=\dfrac{3}{2}\left( {{\text{V}}_{0}}\text{(3}{{\text{P}}_{\text{0}}}\text{-}{{\text{P}}_{\text{0}}}) \right)$
$\therefore \text{ Q}=\text{3}{{\text{P}}_{0}}{{\text{V}}_{0}}$ ….equation (1)
This the heat absorbed from A to B.
The path from B to C is called a isobaric process, since the pressure remains constant in this process, the heat absorbed in this process is given by the equation,
$\text{Q}=\text{n}{{\text{C}}_{\text{P}}}\Delta T$
Where, ${{\text{C}}_{\text{V}}}$ is the specific heat of a monoatomic gas at constant volume, whose value is $\dfrac{5}{2}\text{R}$.
$\therefore \text{ Q}=\dfrac{5}{2}(\text{nR}\Delta T)$
We know that from the ideal gas equation, PV=nRT, so substituting this in above equation we get,
$\text{Q}=\dfrac{5}{2}\left( \text{P}\Delta V \right)$
$\text{Q}=\dfrac{5}{2}\left( \text{3}{{\text{P}}_{0}}\left( \text{2}{{\text{V}}_{\text{0}}}\text{-}{{\text{V}}_{\text{0}}} \right) \right)$
$\therefore \text{Q}=\dfrac{15}{2}{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}$ ……. Equation (2)
This is the heat absorbed in the path B to C.
The total work done by the system is the sum of the work done in the paths B to C and D to A.
The work done in the path from B to C is given by the formula, $\text{W}=\text{P }\\!\\!\Delta\\!\\!\text{ V}$, so the work done can be written as
${{\text{W}}_{1}}=\text{3}{{\text{P}}_{\text{0}}}\text{(2}{{\text{V}}_{\text{0}}}\text{-}{{\text{V}}_{\text{0}}})$
$\therefore {{\text{W}}_{1}}=3{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}$ …. Equation (4)

The work done in the path from C to D, is given by the formula, $\text{W}=\text{P }\\!\\!\Delta\\!\\!\text{ V}$, so the work done can be written as
${{\text{W}}_{2}}={{\text{P}}_{\text{0}}}\text{(}{{\text{V}}_{\text{0}}}\text{-2}{{\text{V}}_{\text{0}}})$
$\therefore {{\text{W}}_{2}}=-{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}$ …. Equation (5)
From equations (4) and (5), we have the total work done as,
$\text{W}={{\text{W}}_{\text{1}}}+{{\text{W}}_{\text{2}}}$
$\text{W}=\text{3}{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}-{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}$
$\therefore \text{W}=\text{2}{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}$ ……equation (6)
Therefore the efficiency $\left( \text{ }\\!\\!\eta\\!\\!\text{ } \right)$ of the cycle is given by the formula,
$\text{ }\\!\\!\eta\\!\\!\text{ }=\dfrac{\text{Work Done}}{\text{Heat Absorbed}}$
From equations (1),(2) and (6), we have
$\text{ }\\!\\!\eta\\!\\!\text{ }=\dfrac{\text{W}}{3{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}\text{+}\dfrac{\text{15}}{\text{2}}{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}}$
$\text{ }\\!\\!\eta\\!\\!\text{ }=\dfrac{2{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}}{\dfrac{21}{2}{{\text{P}}_{\text{0}}}{{\text{V}}_{\text{0}}}}$
$\therefore \text{ }\eta =\dfrac{4}{21}$
So the efficiency of the cycle is $\dfrac{4}{21}$.
So the answer to the question is option (A).

Note: Isothermal process is a process in which the temperature remains constant throughout the process. Since there is no temperature change, there will be no internal energy change as internal energy is temperature dependent.
If you know the specific heat capacity at constant volume or pressure of any gas, we can calculate the other specific heat at constant pressure or volume by using the relation, ${{\text{C}}_{\text{P}}}-{{\text{C}}_{\text{V}}}=\text{R}$, where R is the gas constant.