Question

Two liters of atm and 5 atm are expanded isothermally against a constant external pressure of 1 atm until the pressure of the gas reaches 1 atm. Assuring the gas to be ideal. Calculate the work of expansion.
(A) 400J
(B) -810J
(C) -810KJ
(D) 600KJ

Answer 1

Hint: In order to calculate the work of expansion, we must know about Boyle's law. We can solve this problem only by using Boyle's law. Boyle’s law is said to be a type of gas law which states that the pressure which is exerted by the gas at constant temperature is inversely proportional to the volume occupied by the gas.

Complete step by step answer:
- Let us first understand Boyle's law. Boyle’s law is said to be a type of gas law which states that the pressure which is exerted by the gas at constant temperature is inversely proportional to the volume occupied by the gas. In other words, we can say that as long as the temperature and the quantity of the gas kept constant, then the pressure and the temperature is inversely proportional to each other. The relationship is expressed as follows:
\[P\propto \dfrac{1}{V}\]
- Any change in the volume occupied by the gas will also show change in the pressure exerted by the gas.
Therefore, the Boyle’s law is expressed as
\[{P_1}{V_1} = {P_2}{V_2}\]
- The work of expansion can be determined by using the following formula
\[w = - {P_{ext}}({V_2} - {V_1})\]
\[w = - 1({V_2} - {V_1})\]
Given,
\[{V_1}\] = 2 litres
\[{V_2}\] = ?
\[{P_1}\]= 5 atm
\[{P_2}\]= 1 atm
T= 273K
\[{P_1}{V_1} = {P_2}{V_2}\]
\[{V_2} = \dfrac{{2 \times 5}}{1} = 10\]
\[w = - 1(10 - 2) = - 8 litre.atm\]
\[w = - \dfrac{{8 \times 1.987}}{{0.0821}} calorie\]
\[w = - \dfrac{{8 \times 1.987 \times 4.184}}{{0.0821}} joule\]
\[w = - 810 joule\]
The correct answer is option “B” .

Note: We have to remember that there are different gas laws present other than Boyle’s law:
 - Charle’s law
- Ideal gas law
- Gay-Lussac’s law
- Combined gas law
- Dalton's law of partial pressure