Question

A weather balloon filled with hydrogen at 1 atm and $\text{2}{{\text{7}}^{\text{0}}}\text{C}$ has volume equal to \[\text{12000 litres}\]. On ascending it roaches a place where the temperature is $-\text{2}{{\text{3}}^{\text{0}}}\text{C}$ and pressure is$\text{0}\text{.5 atm}$. The volume of the balloon is
A) \[\text{24000 litres}\]
B) \[\text{20000 litres}\]
C) \[\text{10000 litres}\]
D) \[\text{12000 litres}\]

Answer 1

Hint: The combined gas law states the relation between Boyle’s, Charles, and Gay-Lussac’s law. According to which the ratio of the product of pressure and temperature to the absolute temperature in kelvin is constant. The T, P, and V of the gas changes with the altitude thus apply the combined gas law to obtain the unknown value.
\[\dfrac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}}\text{ = }\dfrac{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{T}}_{\text{2}}}}\]

Complete step by step solution:
We know that different laws state the relation between the pressure, volume, temperature, and the number of moles of gas. The combined gas law is a law that combines the three gas laws: Boyle’s law, Charles’ law, and Gay-Lussac’s law. According to the combined gas law, the ratio of the product of pressure (P) and volume (V) at the absolute temperature (T) is always equal to constant. It is applicable when the system changes pressure, temperature, and volume. The common formula for the combined gas law which is used to relate the before and after situation of gas is as:
\[\text{ }\dfrac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}}\text{ = }\dfrac{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{T}}_{\text{2}}}}\text{ or }\dfrac{{{\text{P}}_{\text{Initial}}}{{\text{V}}_{\text{Initial}}}}{{{\text{T}}_{\text{Initial}}}}\text{ = }\dfrac{{{\text{P}}_{\text{Final}}}{{\text{V}}_{\text{Final}}}}{{{\text{T}}_{\text{Final}}}}\text{ }\]
In the problem, the weather balloon is filled with $\text{ }{{\text{H}}_{\text{2}}}$gas and allowed to move upward. The given data is as follows:
$\begin{align}
 & {{\text{P}}_{\text{1}}}\text{=1 atm } \\
 & {{\text{T}}_{\text{1}}}\text{ = 2}{{\text{7}}^{\text{0}}}\text{C = 2}{{\text{7}}^{\text{0}}}\text{C + 273K= 300 K} \\
& {{\text{V}}_{\text{1}}}\text{= 12000 Litres} \\
& {{\text{P}}_{\text{2}}}\text{= 0}\text{.5 atm } \\
 & {{\text{T}}_{\text{1}}}\text{ = }-\text{2}{{\text{3}}^{\text{0}}}\text{C = }-\text{2}{{\text{3}}^{\text{0}}}\text{C + 273K= 250 K} \\
 & {{\text{V}}_{\text{1}}}\text{=To find} \\
\end{align}$
Let us apply the combined gas law.
\[\dfrac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}}\text{ = }\dfrac{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{T}}_{\text{2}}}}\]
Substitute the values for the pressure, temperature, and volume in the combined gas law. we get,
\[\begin{align}
  & \dfrac{\left( 1\text{ atm} \right)\left( 12000\text{ liter} \right)}{\left( 300\text{ K} \right)}\text{ = }\dfrac{\left( 0.5\text{ atm} \right){{\text{V}}_{\text{2}}}}{\left( 250\text{ K} \right)} \\
 & {{\text{V}}_{\text{2}}}=\text{ }\dfrac{\left( 1\text{ atm} \right)\left( 12000\text{ liter} \right)\left( 250\text{ K} \right)}{\left( 300\text{ K} \right)\left( 0.5\text{ atm} \right)} \\
 & {{\text{V}}_{\text{2}}}=\text{ }\dfrac{\left( 1\text{ at} \right)\left( 12000\text{ liter} \right)\left( 250\text{ } \right)}{\left( 300\text{ } \right)\left( 0.5\text{ at} \right)} \\
 & {{\text{V}}_{\text{2}}}=\text{ }\dfrac{\left( 3000000 \right)}{\left( 150 \right)}\text{liter} \\
 & {{\text{V}}_{\text{2}}}=\text{ 20000 liter} \\
\end{align}\]
Therefore the volume of the gas at the $\text{-2}{{\text{3}}^{\text{0}}}\text{C}$ temperature and $\text{0}\text{.5 atm}$is\[\text{20000 liter}\].

Hence, (B) is the correct option.

Note: Such a question can be asked at the STP condition. Such problems appear a lot thus memorize it.The STP stands for the “standard temperature and pressure” which is $273\text{ K}$ and $760\text{ mm of Hg}$ or $1\text{ atm}$ pressure.