Question

For $2.8g$ of ${N_2}$ gas at $300K$ and $20\;atm$ was allowed to expand isothermally against a constant external pressure of $1\;atm$. Calculate $W$ for the gas.
(A) $ + 236.95\;J$
(B) $ + 136.95\;J$
(C) $ - 236.95\;J$
(D) $ - 136.95\;J$

Answer 1

Hint: The concepts we will use to solve this numerical problem are the mole concept and ideal gas laws. Isothermal here signifies that the temperature was maintained to be a constant during the experiment under consideration. We substitute the known values in the equations and find out the unknown values to finally find the work done (energy) for the given gas.

Complete answer: Let us initially write down the given values:
The mass of the gas given that is mass of ${N_2}$ let it be denoted as $m$; $m = 2.8g$
The external pressure is given to be ${P_{ext}} = 1\;atm$, this is also the final pressure (${P_2}$)
The original pressure at which expansion took place ${P_1} = 20\;atm$
The constant temperature is provided as $T = 300K$
Keep in mind the ideal gas constant $R = 8.314$
Now let us proceed to finding the value of work done $W = ?$
Now we also know that the molar mass of nitrogen (${N_2}$), let it be $M$; $M = 28g$
By the mole concept the number of moles can be computed using the equation:
$ \Rightarrow n = \dfrac{m}{M}$, where $n = $ number of moles
Substituting the mass and molar mass of nitrogen:
$ \Rightarrow n = \dfrac{{2.8g}}{{28g}}$
$ \Rightarrow n = 0.1\;mole$
This gas is defined to have been expanded in isothermal conditions, this implies that there had been no energy variations ($\Delta U = 0$). This allows us to also classify that the work done by the given gas is equal to the heat energy the gas absorbed.
So we can define the formula for work done by nitrogen gas as:
$ \Rightarrow W = - {P_{ext}} \times \Delta V$
But we do not know the change in volume directly so we use the ideal gas equation to say that volume is:
$ \Rightarrow V = \dfrac{{nRT}}{P}$
Now:
$ \Rightarrow W = - {P_{ext}} \times ({V_2} - {V_1})$
$ \Rightarrow W = - {P_{ext}} \times (\dfrac{{nRT}}{{{P_1}}} - \dfrac{{nRT}}{{{P_2}}})$; this is because the temperature remains constant as it is isothermal.
Now we try to simplify:
$ \Rightarrow W = - {P_{ext}} \times nRT \times (\dfrac{1}{{{P_1}}} - \dfrac{1}{{{P_2}}})$
Substituting the values that we have:
$ \Rightarrow W = - 1 \times 0.1 \times 8.314 \times 300 \times \left( {\dfrac{1}{{01}} - \dfrac{1}{{20}}} \right)$
Solving we get:
$ \Rightarrow W = - 249.42 \times \left( {\dfrac{{19}}{{20}}} \right)$
Simplifying we get:
$ \Rightarrow W = - 236.95\;J$
Clearly after computation we have found the required value of work done to be $W = - 236.95\;J$.
So it is evident that the options: (A) $ + 236.95\;J$, (B) $ + 136.95\;J$ and (D) $ - 136.95\;J$ are incorrect options since they do not hold the computed value of work done by the gas. The option (C) is the only option that holds the correct value of $ - 236.95\;J$.
Therefore clearly the correct answer is in option (C) $ - 236.95\;J$.

Note:
Similar to isothermal processes we have isobaric processes. Isobaric processes are thermodynamic processes that happen at a condition where the pressure remains constant. This effectively cancels out any pressure changes caused by heat transfer. When heat is transmitted to the device in an isobaric phase, there is some quantity of work being carried out.