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8 November, 20:04

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

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Answers (1)
  1. 8 November, 21:41
    0
    5.54 m³.

    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.
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