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ANSWER BOTH PLEASE
1.
A helium-filled weather balloon has a volume of 674 L at 20.9 °C and 759 mmHg. It is
released and rises to an altitude of 6.49 km, where the pressure is 403 mmHg and the
temperature is –21.1 °C.
The volume of the balloon at this altitude is ____L.
2.
A sample of xenon gas occupies a volume of 8.83 L at 50.0 °C and 0.940 atm.
If it is desired to increase the volume of the gas sample to 11.6 L, while increasing its pressure
to 1.16 atm, the temperature of the gas sample at the new volume and pressure must be
_____°C.
Solution
1)
Given:
Pi = 759 mmHg
Pf = 403 mmHg
Vi = 674 L
Ti = 20.9 oC
= (20.9+273) K
= 293.9 K
Tf = -21.1 oC
= (-21.1+273) K
= 251.9 K
use:
(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)
(759.0 mmHg*674.0 L)/(293.9 K) = (403.0 mmHg*Vf)/(251.9 K)
Vf = 1.09*10^3 L
Answer: 1.09*10^3 L
2)
Given:
Pi = 0.940 atm
Pf = 1.16 atm
Vi = 8.83 L
Vf = 11.6 L
Ti = 50.0 oC
= (50.0+273) K
= 323 K
use:
(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)
(0.94 atm*8.83 L)/(323.0 K) = (1.16 atm*11.6 L)/(Tf)
Tf = 524 K
Tf = (524-273) oC
Tf = 251 oC
Answer: 251 oC

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ANSWER BOTH PLEASE 1- A helium-filled weather balloon has a volume of.docx

  • 1. ANSWER BOTH PLEASE 1. A helium-filled weather balloon has a volume of 674 L at 20.9 °C and 759 mmHg. It is released and rises to an altitude of 6.49 km, where the pressure is 403 mmHg and the temperature is –21.1 °C. The volume of the balloon at this altitude is ____L. 2. A sample of xenon gas occupies a volume of 8.83 L at 50.0 °C and 0.940 atm. If it is desired to increase the volume of the gas sample to 11.6 L, while increasing its pressure to 1.16 atm, the temperature of the gas sample at the new volume and pressure must be _____°C. Solution 1) Given: Pi = 759 mmHg Pf = 403 mmHg Vi = 674 L Ti = 20.9 oC = (20.9+273) K = 293.9 K
  • 2. Tf = -21.1 oC = (-21.1+273) K = 251.9 K use: (Pi*Vi)/(Ti) = (Pf*Vf)/(Tf) (759.0 mmHg*674.0 L)/(293.9 K) = (403.0 mmHg*Vf)/(251.9 K) Vf = 1.09*10^3 L Answer: 1.09*10^3 L 2) Given: Pi = 0.940 atm Pf = 1.16 atm Vi = 8.83 L Vf = 11.6 L Ti = 50.0 oC = (50.0+273) K = 323 K use: (Pi*Vi)/(Ti) = (Pf*Vf)/(Tf) (0.94 atm*8.83 L)/(323.0 K) = (1.16 atm*11.6 L)/(Tf) Tf = 524 K Tf = (524-273) oC Tf = 251 oC