Question

A sample of gas at 1.2 atm and 27$^0C$ is heated at constant pressure to 57$^0C$. Its final volume is found to be 4.75 liters. What was its original volume?
(A)- 4.32 liters
(B)- 5.02 liters
(C)- 4.22 liters
(D)- None of these

Answer 1

The above question is based on gas law. The different gas laws deal with how gases behave with respect to pressure, volume, temperature, and amount. The formula to be used is the one when pressure is constant i.e. $\dfrac{V}{T}$= constant where V is the volume and T is temperature.

Complete Solution :
Gas law has various relations of Temperature, pressure, number of moles and volume. Let's talk about different types of gas laws and find out which one is suitable for us to find the solution of the given question.
- Charles’s Law (The Temperature-Volume law) : This law states that when the pressure on a sample of a dry gas is held constant, the temperature (Kelvin) and the volume will be in direct proportion i.e.
$V \propto T$
Another way to describe it is by saying that their ratio (V/T) are constant.
$\dfrac{V}{T}$ = Constant
When the Volume increases, Temperature also increases and vice versa.

- Boyle's Law (The Pressure-Volume Law): Boyle's law also known as pressure-volume law states that the volume of a given amount of gas varies inversely with the applied pressure when the temperature and mass are constant.
$V \propto \dfrac{1}{P}$
Another way to describe it is by saying that their products (PV) are constant.
PV = Constant
When pressure goes up, at the same time volume goes down. When volume goes up, pressure goes down i.e. vice-versa.
From the equation above, this can be derived:
${P_1}{V_{_1}} = {P_2}{V_2} = {P_3}{V_3}$

- Gay-Lussac's Law (The Pressure Temperature Law): This law states that the pressure of a given amount of gas at constant volume is directly proportional to the temperature (in Kelvin).
$P \propto T$
Same as before, a constant can be put in so the equation becomes as follows:
$\dfrac{P}{T}$ = Constant
With an increase in temperature, the pressure will go up and vice-versa.
Same as before, initial and final pressures and temperatures (under constant volume and constant number of moles) can be calculated.

- Avogadro's Law (The Volume Amount Law): This law gives us the relationship between volume and amount of gas when pressure and temperature are held constant. Remember that amount is measured in the number of moles. Also, since volume is one of the variables, that means the container holding the gas can somehow expand or contract.
If the amount of gas in a container increases, its volume also increases. If the amount of gas in a container is decreased, the volume decreases too.
$V \propto n$
As before, a constant can be put in, so the equation becomes:
$\dfrac{V}{n}$ = Constant
This means that the fraction of volume and number of moles will always be the same value if the pressure and temperature remain constant.

- The Combined Gas Law: Now we can combine everything we have discussed till now into one proportion, that is:
$V \propto \dfrac{T}{P}$
The volume of a given amount of gas is proportional to the ratio of its temperature (Kelvin) and its pressure at a constant number of moles.
Same as before, a constant can be put in. So, the equation becomes:
$\dfrac{{PV}}{T} = C$
As the pressure goes up, the temperature also goes up, and vice-versa.

- The ideal gas law : According to Avogadro, the same volumes of gas contain the same number of moles, chemists could now determine the formula of gaseous elements and their formula masses and by combining all the above formulas, we not got a formula of formulas known as the ideal gas law and it can be written as :
$PV = nRT$
Where n is the number of moles of the number of moles and R is a constant called the universal gas constant
According to the conditions given in question, the one which is useful for us to find our answer is Charle's Law. By applying Charle's law, we get
$\begin{gathered} \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \\\ \dfrac{{{V_1}}}{{300}} = \dfrac{{4.75}}{{330}} \\\ {V_1} = 4.32{\text{ litres}} \\\ \end{gathered}$
So, the correct answer is “Option A”.

Note: Always remember that all the above formula is only for ideal gases and there are many assumptions made before defining these laws. For real gases, we use a term compressibility factor, denoted by Z, to convert the ideal gas law equation to use for real gasses too.