Question

A balloon blown up with 1 mole of a gas, has a volume of 480 ml to \[\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}\]C. If the balloon is filled to ${{\dfrac{7}{8}}^{th}}$of its maximum capacity, then which of the following options is/are correct?
A. The balloon will burst at ${{30}^{{}^\circ }}$C.
B. The pressure of the gas inside the balloon at ${{5}^{{}^\circ }}$C is 47.5 atm.
C. The minimum temperature at which the balloon will burst is \[{{44.71}^{{}^\circ }}\]C
D. The pressure of gas when balloons burst at minimum temperature is 50 atm.

Answer 1

Hint: The concept is basically based on Charles law which is also defined as law of volumes which gives us explanation about how gas expands with the increase in temperature and the converse is also true that decrease in temperature will lead to decrease in volume.

Complete step-by-step answer:
The main equation given by Charles law can be written as
\[\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}\]
Maximum capacity i.e. volume of balloon = ${{\dfrac{8}{7}}^{th}}$of whole volume i.e. $\dfrac{8}{7}\times 480=548.57ml$
The value of ${{V}_{1}}$= 480 ml, ${{T}_{1}}$= 278 K, n = 1 mole (Given)
Now by using Charles law we know the equation can also be written as: \[\dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}}\], now by putting the values in this equation
\[\dfrac{480}{278}=\dfrac{548.57}{{{T}_{2}}}\]
Therefore the value of \[{{T}_{2}}\]= 317.71 K or we can say that it is equal to \[{{44.71}^{{}^\circ }}C\].

From this calculation we can consider that the minimum temperature at which the balloon will burst is \[{{44.71}^{{}^\circ }}\]C, option C is the correct answer.

Note: Charles law is just the special case of ideal gas law and the ideal gas is PV = nRT where P is the pressure measured in atmospheres, V is volume, n is number of moles of gas, R is constant and T is temperature measured in kelvin.