The combined gas law: A.) a very flexible helium-filled balloon is released from
ID: 903820 • Letter: T
Question
The combined gas law:A.) a very flexible helium-filled balloon is released from the ground into the air at 20 degrees Celcius. The initial volume of the balloon is 5.00 L, and the pressure is 716 mmHg. The balloon I send to an altitude of 20 KM, where the pressure is 76.0 mmHg and the temperature is -50 degrees Celcius. What is the volume, V2, of the balloon in liters, assuming it doesn't break or leak?
V2= ? L
B.) consider 4.00 L of gas at 365 mmHg and 20 degrees Celcius. Is there a container is compress to 2.20 L and the temperature is increased to 39°C what is the new pressure, inside the container? I assume no change in the amount of gas inside the cylinder
P2= ? MmHg The combined gas law:
A.) a very flexible helium-filled balloon is released from the ground into the air at 20 degrees Celcius. The initial volume of the balloon is 5.00 L, and the pressure is 716 mmHg. The balloon I send to an altitude of 20 KM, where the pressure is 76.0 mmHg and the temperature is -50 degrees Celcius. What is the volume, V2, of the balloon in liters, assuming it doesn't break or leak?
V2= ? L
B.) consider 4.00 L of gas at 365 mmHg and 20 degrees Celcius. Is there a container is compress to 2.20 L and the temperature is increased to 39°C what is the new pressure, inside the container? I assume no change in the amount of gas inside the cylinder
P2= ? MmHg
A.) a very flexible helium-filled balloon is released from the ground into the air at 20 degrees Celcius. The initial volume of the balloon is 5.00 L, and the pressure is 716 mmHg. The balloon I send to an altitude of 20 KM, where the pressure is 76.0 mmHg and the temperature is -50 degrees Celcius. What is the volume, V2, of the balloon in liters, assuming it doesn't break or leak?
V2= ? L
B.) consider 4.00 L of gas at 365 mmHg and 20 degrees Celcius. Is there a container is compress to 2.20 L and the temperature is increased to 39°C what is the new pressure, inside the container? I assume no change in the amount of gas inside the cylinder
P2= ? MmHg
Explanation / Answer
A)
T1 = 20 oC = (20+273) K = 293 K
V1= 5L
P1 = 716 mm
P2=76 mm
T2= -50 oC = ( -50+273) K = 223 K
V2=?
use:
P1*V1/T1 = P2*V2/T2
716*5/293 = 76*V2/223
V2= 35.9 L
Answer: 35.9 L
B)
T1 = 20 oC = (20+273) K = 293 K
V1= 4L
P1 = 365 mm
T2= 39 oC = ( 39+273) K = 312 K
V2= 2.2 L
use:
P1*V1/T1 = P2*V2/T2
365*4/293 = P2*2.2/312
P2=706.7 mm Hg
Answer: 706.7 mm Hg