The Combined Gas Law All Pressure-volume-temperature relationships for gasses ca

Question

The Combined Gas Law All Pressure-volume-temperature relationships for gasses can be combined into a single relationship known as the combined gas law. This expression can be used when looking at the effect of changes in two of these variables on the third as long as the amount of gas (number of moles) remains constant. To use the combined gas law property. You must always express the temperatures in kelvins. The combined gas law can be represented as follows: P_1V_1/T_1 = P_2V_2/T_2 Avery flexible helium-filled balloon is released from the ground into the air at 20. degree C. The initial volume the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an attitude 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 Consider 4.60 L of a gas at 365 mmHg and 20. degree C, if the container is compressed to 2.80 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.

Explanation / Answer

A)

T1 = 20°C = 293 K

V1 = 5 L, P1 = 760 mm Hg

P2 = 76 mm Hg, T2 = -50° = 223K V2 = ?

apply idal gas law

P1V1/T1 = P2V2/T2

solve for V2

V2 = P1V1/T1*T2/P2

V2 = 760/76 * 223/293 * 5 = 38.05 L

B)

V1 = 4.6 L , P1 = 365 mm Hg, T1 = 20°C = 293

V2 = 2.8 L, P2 = x ? . T2 30|C = 303 K

substitute

P1V1/T1 = P2V2/T2

P2 = P1*(V1/V2)(T2/T1)

P2 = 365 * 4.6/2.8*303/293

P2 = 620.10 mm Hg

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