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, *V*2, 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, *P*2, inside the container? Assume
no change in the amount of gas inside the cylinder.

Answer #1

a) Let; We have given;

P_{1}= 760mmHg = **1 atm**

T_{1} = 20^{o}C = 20+273.15=
**293.15K**

V_{1} = **5.00L**

**Also;** P_{2} =76.0mmHg
=**0.1atm**

** **T_{2} =-50^{o}C =
-50 +273.15 =**223.15K**

**V _{2} = ? To calculate.**

**We know; Combined gas law;**

(P_{1} x V_{1})/T_{1} =(P_{2} x
V_{2})/T_{2}

V_{2} =(P_{1} x V_{1} x
T_{2})/(T_{1} x P_{2})

Putting the values in the formula;

V_{2} = (1atm x5.00Lx223.15K)/(293.15K x0.1atm)

**V _{2} = 38.06L**

b) Let; We have given;

P_{1}= 365mmHg = **0.4802 atm**

T_{1} = 20^{o}C = 20+273.15=
**293.15K**

V_{1} = **4.20L**

**Also;** P_{2} =? **To
calculate.**

** **T_{2} =32^{o}C = 32
+273.15 =**305.15K**

V_{2} = **2.80L**

**We know; Combined gas law;**

(P_{1} x V_{1})/T_{1} =(P_{2} x
V_{2})/T_{2}

P_{2} =(P_{1} x V_{1} x
T_{2})/(T_{1} x V_{2})

Putting the values in the formula;

P_{2}= (0.4802atm x4.20Lx305.15K)/(293.15K x2.80L)

**P _{2} = 0.7498 atm**

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