Question

A balloon is filled with hydrogen at room temperature, it will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, up to what volume can the balloon be expanded?

Answer 1

Hint: Boyle's law is an experimental gas law that tells us how the pressure of a gas tends to increase as there is a decrease in the volume of the container at constant temperature.
\[{{P}_{1}}{{V}_{1}}\,=\,{{P}_{2}}{{V}_{2}}\].

Complete step by step solution:

Boyle observed that the product of the pressure and volume are observed to be nearly constant. The product of pressure and volume is exactly a constant for an ideal gas.
P × V = constant.
It states that for a fixed amount of gas and constant temperature the volume of a sample of a gas varies inversely with its pressure
P α 1/V or V α 1/P
P = k/V
PV = k
PV = constant
\[{{P}_{1}}{{V}_{1}}\,=\,{{P}_{2}}{{V}_{2}}\]
Where, \[{{P}_{1}}\] and \[{{V}_{1}}\] are the initial pressure and volume and \[{{P}_{2}}\] and \[{{V}_{2}}\] are the values of the pressure and volume of the gas after change.
Now, in the given question the initial pressure (\[{{P}_{1}}\]) is 1 bar, the balloon will burst if the pressure exceeds 0.2 bar, that means the final pressure (\[{{P}_{2}}\]) will be 0.2 bar.
Similarly, the initial volume (\[{{V}_{1}}\]) is given i.e. 2.27 L and we have to find the final volume (\[{{V}_{2}}\]).
By applying Boyle’s law
\[{{P}_{1}}{{V}_{1}}\,=\,{{P}_{2}}{{V}_{2}}\]
1 × 2.27 = 0.2 × \[{{V}_{2}}\]
\[{{V}_{2}}\] = 1 × 2.27/ 0.2
\[{{V}_{2}}\]= 11.35 L
Therefore, from the above solution we can conclude that the volume till which the balloon can be expanded is 11.35 L.

Note: If we represent Boyle's law graphically, the Graph between P and V at constant temperature is called isotherm and is an equilateral hyperbola.
At constant mass and temperature density of a gas is directly proportional to its pressure and inversely proportional to its volume.