A weather balloon inflated with helium has a volume of 5.0\text{ dm}^3 at ground level whe… — JAMB UTME

A weather balloon inflated with helium has a volume of $5.0\text{ dm}^3$ at ground level where the atmospheric pressure is $100\text{ kPa}$ and the temperature is $27^\circ\text{C}$ . The balloon ascends to an altitude where the ambient pressure drops to $40\text{ kPa}$ . Assuming isothermal conditions are maintained throughout the ascent, what is the new volume of the helium gas in the balloon?

$12.5\text{ dm}^3$ Answer
B
$2.0\text{ dm}^3$
C
$1.125\text{ dm}^3$
D
$8.0\text{ dm}^3$

Answer

12.5\text{ dm}^3

Boyle's Law states that for a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure (

P_1 V_1 = P_2 V_2

). Substituting the initial pressure of

100\text{ kPa}

, initial volume of

5.0\text{ dm}^3

, and final pressure of

40\text{ kPa}

gives

V_2 = (100 \times 5.0) / 40 = 12.5\text{ dm}^3

Step-by-Step Solution

Identify the given parameters and constant conditions

P_1 = 100\text{ kPa}

V_1 = 5.0\text{ dm}^3

P_2 = 40\text{ kPa}

. Temperature is constant (isothermal condition).

Boyle's Law applies when a fixed mass of gas is kept at a constant temperature.

State Boyle's Law formula

P_1 V_1 = P_2 V_2

Pressure and volume of a fixed mass of ideal gas are inversely proportional at constant temperature.

Rearrange the formula to solve for the unknown final volume (

V_2

)

V_2 = \frac{P_1 V_1}{P_2}

Isolating

V_2

allows direct substitution of known variables.

Substitute the given numerical values into the equation and calculate

V_2 = \frac{100\text{ kPa} \times 5.0\text{ dm}^3}{40\text{ kPa}} = \frac{500}{40} = 12.5\text{ dm}^3

Units of pressure (

\text{kPa}

) cancel out, leaving the volume in

\text{dm}^3

Key Concept

Boyle's Law (

P_1 V_1 = P_2 V_2

at constant temperature)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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