Question

Before it was launched, a helium-filled balloon had a pressure of 101 kPa at a temperature of 20°C. At an altitude of 15 000 m, the pressure had decreased to 11.7 kPa and the temperature had dropped to -56 °C. The volume of the balloon increased to 35. 4 m3.

What is the original volume of the ballon
6.9 m3
1.34 m3
2.56 m3
43.0 m3
5.54 m3

Answer 1

5.54 m³.

Step-by-step explanation:

• We should use the ideal gas law: PV = nRT,

where, P is the pressure of the gas in atm,

V is the volume of the gas in L,

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K.

We have two different cases at different (P, V and T) while the number of moles of He and R is constants.

∴ (P₁V₁) / (T₁) = (P₂V₂) / (T₂).

We can use P in KPa and V in m³ that the conversion factor can be canceled by division, but we should use T in K because its conversion factor is additive value.

P₁ = 101.0 kPa, V₁ = ??? m³, and T₁ = (20 °C + 273) = 293.0 K.

P₂ = 11.7 kPa, V₂ = 35.4 m³, and T₂ = (-56 °C + 273) = 217 K.

∴ The initial volume (V₁) = (P₂V₂T₁) / (P₁T₂) = (11.7 kPa)(35.4 m³)(293.0 K) / (101.0 kPa)(217.0 K) = (121354.74) / (21917) = 5.537 m³ ≅ 5.54 m³.