Question

Calculate the work done by the system in an irreversible (single step) adiabatic expansion of $$2$$ moles of a polyatomic gas $$\left(\gamma ={\displaystyle \frac{4}{3}}\right)$$ from $$300\text{K}$$ and pressure $$10\text{atm}$$ to $$1\text{atm}$$ :
(A) $$-227\text{R}$$
(B) $$-787.8\text{R}$$
(C) $$-202.5\text{R}$$
(D) $$-3.367\text{kJ}\text{mol}{\text{e}}^{-1}{\text{K}}^{-1}$$

Answer 1

Hint:First of all, we will write a relation involving temperature and pressure for an adiabatic process, then will equate for the initial and the final phases to find the final temperature. Using the appropriate formula, we will find the work done in terms of the gas constant.

Complete step by step solution:
In the given question, we are supplied the following data:
Initial temperature is $$300\text{K}$$ .
Initial pressure is $$10\text{atm}$$ .
Final pressure is $$1\text{atm}$$ .
Specific heat ratio is given as $${\displaystyle \frac{4}{3}}$$ .
We are asked to calculate the work done by the system in an irreversible (single step) adiabatic expansion.
Number of moles of gas is given as $$2$$ moles.
The gas is polyatomic in nature.
To begin with, we will need to find the final temperature, so that we can find the change in temperature so that we can calculate the work done.Let us proceed to solve the problem.We know, for an adiabatic process, we have a relation which is given below:
$${\displaystyle \frac{T^{\gamma }}{P^{\gamma -1}}}=\text{c}$$ …… (1)
Where,
$$T$$ indicates the temperature of the gas.
$$P$$ indicates the pressure of the gas.
$$\gamma$$ indicates the specific heat ratio of the gas.
$$\text{c}$$ indicates constant.
So, we can write for the gas in the initial temperature and pressure:
$${\displaystyle \frac{T_1^{\gamma }}{P_1^{\gamma -1}}}=\text{c}$$ …… (2)
And, for the final temperature and pressure, we can write:
$${\displaystyle \frac{T_2^{\gamma }}{P_2^{\gamma -1}}}=\text{c}$$ …… (3)
Now, we divide the equation (2) and (3), and we get:
$$\begin{gathered} {\displaystyle \frac{\left({\displaystyle \frac{T_1^{\gamma }}{P_1^{\gamma -1}}}\right)}{\left({\displaystyle \frac{T_2^{\gamma }}{P_2^{\gamma -1}}}\right)}}=1 \\ \Rightarrow {\displaystyle \frac{T_1^{\gamma }}{P_1^{\gamma -1}}}={\displaystyle \frac{T_2^{\gamma }}{P_2^{\gamma -1}}} \\ \Rightarrow {\displaystyle \frac{T_1^{\gamma }}{T_2^{\gamma }}}={\displaystyle \frac{P_1^{\gamma -1}}{P_2^{\gamma -1}}} \end{gathered}$$
$$\Rightarrow {\left({\displaystyle \frac{T_1}{T_2}}\right)}^{\gamma }={\left({\displaystyle \frac{P_1}{P_2}}\right)}^{\gamma -1}$$ …… (4)
Now, we substitute the required values in the equation (4) and we get:
$$\begin{gathered} {\left({\displaystyle \frac{T_1}{T_2}}\right)}^{\gamma }={\left({\displaystyle \frac{P_1}{P_2}}\right)}^{\gamma -1} \\ \Rightarrow {\left({\displaystyle \frac{300}{T_2}}\right)}^{{\displaystyle \frac{4}{3}}}={\left({\displaystyle \frac{10}{1}}\right)}^{{\displaystyle \frac{4}{3}}-1} \\ \Rightarrow {\left({\displaystyle \frac{300}{T_2}}\right)}^{{\displaystyle \frac{4}{3}}}={\left({\displaystyle \frac{10}{1}}\right)}^{{\displaystyle \frac{1}{3}}} \\ \Rightarrow {\displaystyle \frac{300}{T}}={\left(10\right)}^{{\displaystyle \frac{1}{4}}} \end{gathered}$$
Again, we manipulate further and we get:
$$\begin{gathered} T_2={\displaystyle \frac{300}{1.77}} \\ \Rightarrow T_2=168.7\text{K} \end{gathered}$$
Therefore, the final temperature is found to be $$168.7\text{K}$$ .
Now, we calculate the temperature change:
$$\begin{gathered} \mathrm{\Delta }T=T_2-T_1 \\ \Rightarrow \mathrm{\Delta }T=168.7-300 \\ \Rightarrow \mathrm{\Delta }T=-131.3\text{K} \end{gathered}$$
Now, we have the expression for work done for an adiabatic expansion, which is given by:
$$W=n{C}_{\text{v}}\mathrm{\Delta }T$$
$$\Rightarrow W=n\times {\displaystyle \frac{R}{\gamma -1}}\times \mathrm{\Delta }T$$ …… (5)
Where,
$$R$$ indicates the gas constant.
Now, we substitute the required values in the equation (5) and we get:
$$\begin{gathered} W=n\times {\displaystyle \frac{R}{\gamma -1}}\times \mathrm{\Delta }T \\ \Rightarrow W=2\times {\displaystyle \frac{R}{{\displaystyle \frac{4}{3}}-1}}\times \left(-131.3\right) \\ \Rightarrow W=6R\times \left(-131.3\right) \\ \therefore W=-787.8\text{R} \end{gathered}$$
Hence, the work done by the system in an irreversible (single step) adiabatic expansion is $$-787.8\text{R}$$ .

The correct option is (B).

Note: It is important to remember that in an adiabatic process there is no heat transfer. But the temperature does not remain constant. The magnitude of work done is directly proportional to the temperature change. Work is negative as it is done by the system. Work done on a system is positive in nature.