A very flexible helium-filled balloon is released from the ground into the air a

Question

A very flexible helium-filled balloon is released from the ground into the air at 20. degree C. The initial volume of the balloon is 5.00 L, and the pressure is 760.mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is -50.degree C. What is the new volume, V_2, of the balloon in liters, assuming it doesn't break or leak? Express your answer with the appropriate units. V_2 = Consider 4.80 L of a gas at 365 mmHg and 20 degree C. If the container is compressed to 2.40 L and the temperature is increased to 30. degree C, what is the new pressure, P_2, inside the container? Assume no change in the amount of gas inside the cylinder. Express your answer with the appropriate units. P_2 =

Explanation / Answer

PART(A)

Initial Volume, V1 = 5.00 L

Initial Pressure, P1 = 760. mmHg = 1 atm

Initial Temperature, T1 = 20.0 + 273.15 = 293.15 K

Final Volume, V2 = ?

Final Pressure, P2 = 76.0 mmHg = 0.100 atm

Final Temperature, T2 = - 50.0 + 273.15 = 223.15 K

Useful Formula,

P1V1/T1 = P2V2/T2

Then, V2 = P1 V1 T2 / T1 P2

V2 = (1.00 * 5.00 * 223.15) / (293.15 * 0.100)

V2 = Final volume = 38.0 L

PART(B)

Initial Volume, V1 = 4.80 L

Initial Pressure, P1 = 365 mmHg = 365 / 760 atm = 0.480 atm

Initial Temperature, T1 = 20.0 + 273.15 = 293.15 K

Final Volume, V2 = 2.40 L

Final Pressure, P2 = ?

Final Temperature, T2 = 30.0 + 273.15 = 303.15 K

Useful Formula,

P1V1/T1 = P2V2/T2

Then, P2 = P1 V1 T2 / T1 V2

P2 = (0.480 * 4.80 * 303.15) / (293.15 * 2.40)

P2 = Final Pressure = 0.993 atm

P2 = 0.993 * 760 = 755. mmHg

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