A weather balloon is inflated to a volume of 29.1 L at a pressure of 733 mmHg and a temperature of 24.7 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 365 mmHg and the temperature is - 14.5 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answer: 50.8 L

Explanation: Using the Combined Gas Law:

PV/T = C

Hence: P1*V1/T1 = C

Using the first scenario we will find value of C:

V1 = 29.1 L = 0.0291 m^3

P1 = 733 mmHg = 97725.3 Pa

T1 = 24.7 C = 297.85 K

C = (0.0291 * 97725.3) / 297.85 = 9.56

Use value of C to find V2:

P2 = 365 mmHg = 48662.7 Pa

T2 = - 14.5 C = 258.65 K

C = 9.56

V2 = C*T2/P2 = (9.56 * 258.65) / 48662.7 = 0.0508 m^3

Hence Volume of balloon at higher altitude = 0.0508 m^3 = 50.8 L

volume balloon 5.06 10⁻² m³

Explanation:

The balloon is inflated with air, which in the first instance behaves like an ideal gas, which has as an equation

PV = n R T

Where R is the ideal gas constant that is worth 8.314 J / mol K.

Let's use the initial data to calculate the number of moles of gas in the balloon that remains constant, but first we reduce all quantities to the SI system

V1 = 29.1 L (1 m2 / 1000 L) = 29.1 10⁻³ m³

P1 = 733 mmHg (1 105 Pa / 760 mmHg) = 0.964 10⁵ Pa

T1 = 24.7 + 273.15 = 297.85K

P2 = 365 mmHg (1 105 Pa / 760 mmHg) = 0.480 10⁵ Pa

T2 = - 14.5 + 273.15 = 258.65K

Let's clear and calculate the number of moles

n = P V / R T

n = P1 V1 / R T1

n = 0.964 10 5 29.1 10⁻³ / (8.314 297.85)

n = 1.13 moles

With the data of the second part we clear the volume and calculate

V = n R T / P

V2 = n R T2 / P2

V2 = 1.13 8.314 258.65 / 0.480 10⁵

V2 = 5.062 10⁻² m³