Question

A weather Balloon is inflated with Helium. The Balloon has a volume of $ 100{m^3} $ and it must be inflated to a pressure of $ 0.10atm $ . If $ 50L $ gas cylinders of helium at a pressure of $ 100atm $ are used, How many cylinders are needed? Assume that temperature is constant.
(A) $ 2 $
(B) $ 3 $
(C) $ 4 $
(D) $ 1 $ $ $

Answer 1

Hint: In questions where the temperature is kept constant and no details about the nature of gas is provided we assume Gases to behave ideally, Thus here we can use Boyle's Law which states that for an ideal gas under constant temperature the product of Pressure and Volume is always a constant provided there is no change in number of moles of gas.

Complete step by step solution:
Let us take the total volume of the balloon to be equal to $ {V_1} $ and the final pressure of the balloon to be $ {P_1} $ thus their product equals $ {P_1}{V_1} $ .
Now let us take the pressure of one cylinder to be equal to $ {P_2} $ and volume of one cylinder to be $ {V_2} $ so, For one cylinder the product of pressure and volume equals to $ {P_2}{V_2} $ .
Say we require $ N $ number of cylinders to inflate the balloon completely so the pressure and volume product of $ N $ cylinders is equal to $ N{P_2}{V_2} $ equating this equal to $ {P_1}{V_1} $ we can solve and get the value of $ N $ , However there is a catch which lies in the fact that the volume of the balloon is provided in a different unit and thus it has to be converted into Liters as used for representing cylinders volume so as to get an accurate answer. So in the question we are given:
 $ \begin{array}{*{20}{l}}
  {{P_1} = 0.10{\text{ }}atm} \\
  {{V_1} = 100{m^3} = 100000L} \\
  {{P_2} = 100{\text{ }}atm} \\
  {{V_2} = 50L}
\end{array} $
Now, substituting the values in $ {P_1}{V_1} = N{P_2}{V_2} $ , we get:
 $ 100000 \times 0.1 = N \times 100 \times 50 $
 $\dfrac{{100000 \times 0.1}}{{100 \times 50}} = N \\$
 $N = 2 $
We get $ N = 2 $ , Thus the correct answer is Option (A) i.e. $ 2 $ .

Note:
You must always be careful about units in such numerical problems as a slight error in units can change the answer by a large margin. Also there lies the possibility where cylinders of different volumes and pressure might be used in those cases just sum up the Products. In cases where the nature of gas is provided as a Real gas, Boyle’s law becomes invalid.