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13 January, 19:58

A meteorologist balloon contains 250.0 L of He at 22.0 C and 740.0 mmHg. If the volume of the balloon can vary according to external conditions what volume would it occupy at an altitude at which the temperature is - 52.0 C and the pressure is. 750 atm?

A. 240 L

B. 145 L

C. 760 L

D. 184790 L

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Answers (1)
  1. 13 January, 20:15
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    Answer: Option A) 240 L

    Explanation:

    Given that:

    Initial Volume of helium (V1) = 250.0L

    Initial Temperature of helium (T1) = 22.0°C

    [Convert temperature in celsius to Kelvin by adding 273

    (22.0°C + 273 = 295K) ]

    Initial Pressure of helium (P1) = 740 mmHg

    [convert pressure in mmHg to atmosphere

    If 760 mmHg = 1 atm

    740 mmHg = 740/760 = 0.97 atm]

    Final Volume of helium (V2) = ?

    Final temperature T2 = - 52.0°C

    [Convert - 52°C to Kelvin by adding 273

    -52°C + 273 = 221K]

    Final pressure of helium = 0.750 atm

    Since pressure, volume and temperature are given, apply the combined gas equation

    (P1V1) / T1 = (P2V2) / T2

    (0.97 x 250.0) / 295 = (0.750 x V2) / 221

    242.5/295 = 0.750V2/221

    Cross multiply

    242.5 x 221 = 0.75V2 x 295

    53592.5 = 221.25V2

    V2 = 53592.5 / 221.25

    V2 = 242.2 L (Rounded to the nearest tens, V2 becomes 240 L)

    Thus, the new volume of helium at the altitude is 240 liters
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