Question

A. A very flexible helium-filled balloon is released from the ground into the air at 20. ∘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. ∘C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak? Express your answer with the appropriate units.

B. Consider 4.40 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.10 L and the temperature is increased to 32 ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder. Express your answer with the appropriate units.

Answer #1

A)

Given:

Pi = 760 mmHg

Pf = 76.0 mmHg

Vi = 5.00 L

Ti = 20.0 oC

= (20.0+273) K

= 293 K

Tf = -50.0 oC

= (-50.0+273) K

= 223 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(760 mmHg*5 L)/(293.0 K) = (76 mmHg*Vf)/(223.0 K)

Vf = 38.1 L

Answer: 38.1 L

B)

Given:

Pi = 365 mmHg

Vi = 4.40 L

Vf = 2.10 L

Ti = 20.0 oC

= (20.0+273) K

= 293 K

Tf = 32.0 oC

= (32.0+273) K

= 305 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(365 mmHg*4.4 L)/(293.0 K) = (Pf*2.1 L)/(305.0 K)

Pf = 796 mmHg

= 796/760 atm

= 1.05 atm

Answer: 1.05 atm

All pressure–volume–temperature relationships 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:
P1V1/T1=P2V2/T2
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