Question

# a. A very flexible helium-filled balloon is released from the ground into the air at 20....

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?

b. Consider 4.20

L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.80 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.

a) Let; We have given;

P1= 760mmHg = 1 atm

T1 = 20oC = 20+273.15= 293.15K

V1 = 5.00L

Also; P2 =76.0mmHg =0.1atm

T2 =-50oC = -50 +273.15 =223.15K

V2 = ? To calculate.

We know; Combined gas law;

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

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

Putting the values in the formula;

V2 = (1atm x5.00Lx223.15K)/(293.15K x0.1atm)

V2 = 38.06L

b) Let; We have given;

P1= 365mmHg = 0.4802 atm

T1 = 20oC = 20+273.15= 293.15K

V1 = 4.20L

Also; P2 =? To calculate.

T2 =32oC = 32 +273.15 =305.15K

V2 = 2.80L

We know; Combined gas law;

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

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

Putting the values in the formula;

P2= (0.4802atm x4.20Lx305.15K)/(293.15K x2.80L)

P2 = 0.7498 atm