A weather balloon is launched containing 3.20\text{ dm}^3 of helium gas at an initial temp… — JAMB UTME

A weather balloon is launched containing $3.20\text{ dm}^3$ of helium gas at an initial temperature of $47^\circ\text{C}$ and a pressure of $1.50\text{ atm}$ . As the balloon rises into the atmosphere, the temperature drops to $-23^\circ\text{C}$ and the pressure decreases to $0.60\text{ atm}$ . Calculate the final volume of the helium gas in $\text{dm}^3$ .

Answer: 6.25 dm^3

Answer

The final volume of the helium gas is

6.25\text{ dm}^3

Converting the given temperatures to Kelvin yields

T_1 = 47 + 273 = 320\text{ K}

and

T_2 = -23 + 273 = 250\text{ K}

. Applying the General Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

and rearranging for

V_2

gives

V_2 = \frac{1.50 \times 3.20 \times 250}{0.60 \times 320} = 6.25\text{ dm}^3

Step-by-Step Solution

Convert both initial and final temperatures from Celsius to Kelvin.

T_1 = 47^\circ\text{C} + 273 = 320\text{ K}

and

T_2 = -23^\circ\text{C} + 273 = 250\text{ K}

Gas laws require absolute temperature units (Kelvin).

State the General Gas Law equation.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final values of pressure, volume, and temperature for a fixed mass of gas.

Rearrange the equation to solve for the final volume (

V_2

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}

Isolates the target variable on one side.

Substitute the values into the equation and compute the result.

V_2 = \frac{1.50\text{ atm} \times 3.20\text{ dm}^3 \times 250\text{ K}}{0.60\text{ atm} \times 320\text{ K}} = 6.25\text{ dm}^3

Yields the exact final volume of the gas.

Key Concept

General Gas Law

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