All pressure-volume-temperature relationship for gases can be combined into a si
ID: 539358 • Letter: A
Question
All pressure-volume-temperature relationship for gases 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 properly, 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 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. Consider 4.40 L of a gas at 365 mmHg and 20. degree C. lf the container is compressed to 2.90 L and the temperature is increased to 34 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)
V1 = 5.00 L
P1 = 760 mmHg
T1 = 20 oC = (20 + 273) K = 293 K
V2 = ?
P2 = 76.0 mmHg
T2 = -50 oC = (-50 + 273) K = 223 K
use:
P1*V1/T1 = P2*V2/T2
760*5.00/293 = 76.0*V2/223
V2 = 38.1 L
Answer: 38.1 L
B)
V1 = 4.40 L
P1 = 365 mmHg
T1 = 20 oC = (20 + 273) K = 293 K
V2 = 2.90 L
P2 = ?
T2 = 34 oC = (34 + 273) K = 307 K
use:
P1*V1/T1 = P2*V2/T2
365*4.40/293 = P2*2.90/307
P2 = 580 mmHg
Answer: 580 mmHg