Question

a) A helium-filled weather balloon has a volume of 521 L at 17.9°C and 758 mmHg....

a) A helium-filled weather balloon has a volume of 521 L at 17.9°C and 758 mmHg. It is released and rises to an altitude of 5.22 km, where the pressure is 472 mmHg and the temperature is –14.1°C.

The volume of the balloon at this altitude is ______L

b) A sample of krypton gas occupies a volume of 6.19 L at 56.0°C and 382 torr.

If the volume of the gas sample is decreased to 4.90 L, while its temperature is increased to 129.0°C, the resulting gas pressure will be _____torr

c) A sample of krypton gas occupies a volume of 9.10 L at 42.0°C and 1.29 atm.

If it is desired to increase the volume of the gas sample to 11.5 L, while increasing its pressure to 1.77 atm, the temperature of the gas sample at the new volume and pressure must be ______°C

d) A sample of argon gas at 295 K and 0.514 atm occupies a volume of 2.27 L. If the pressure of the gas is increased, while at the same time it is heated to a higher temperature, the final gas volume _____

°could be larger or smaller than 2.27 L depending on the final pressure and temperature.

°will be smaller than 2.27 L.

°will be larger than 2.27 L.

e) A sample of nitrogen gas at 309 K and 0.150 atm occupies a volume of 2.29 L. If the gas is compressed into a smaller volume, while at the same time it is heated to a higher temperature, the final gas pressure _______
- will be higher than 0.150 atm.

- will be lower than 0.150 atm.

- could be higher or lower than 0.150 atm depending on the final volume and temperature.

f)A sample of hydrogen gas at 320 K and 2.05 atm occupies a volume of 1.50 L. If the gas is allowed to expand to a larger volume , while at the same time it is cooled to a lower temperature, the final gas pressure _______

-could be higher or lower than 2.05 atm depending on the final volume and temperature.

-will be higher than 2.05 atm.

-will be lower than 2.05 atm.

g) Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?
- 10.1 grams of Ne

- 3.01 moles CH4

- 3×1023 molecules of N2

- they would all have the same volume

h) Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?
- 5×1023 molecules of H2

- 16.76 grams of Ne

- 0.831 moles CH4

- they would all have the same volume

i) Which of the following gas samples would have the largest volume, if all samples are at the same temperature and pressure?

- 7×1023 molecules of O2

- 3.01 moles CH4

- 65.7 grams of Xe

- they would all have the same volume

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Answer #1

For all the calculation, we can use following formula

PV= nRT

P = Pressure in atm                       V= Volume in Liter

n = no of moles               R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

Question a

The above formula can be altered as below  

P1V1 / T1 = P2V2 / T2

P1 = 758 mmHg = 0.9973 atm V1 = 521 L T1 = 273 + 17.9 = 290.9 K

P2 =   472 mmHg = 0.621 atm V2 = ?   T2 = 273 + 14.1 = 287.1 K

V2 = 0.9973 atm x 521 L x 287.1 K / 290.9 K x 0.621 atm

V2 = 825.77 Liter

The volume of the balloon at this altitude is 825.77 Liter

Question b

The above formula can be altered as below  

P1V1 / T1 = P2V2 / T2

P1 = 382 Torr = 0.502632 atm V1 = 6.19 L T1 = 273 + 56 = 329 K

P2 = ? V2 = 4.9 Liter T2 = 273 + 129 = 402 K

P2 = 0.502632 atm x 6.19 Liter x 402 K / 4.9 Liter x 329 K

P2 = 0.7758 atm or 589.608 Torr

the resulting gas pressure will be _589.608_torr

Question C

P1V1 / T1 = P2V2 / T2

P1 = 1.29 atm V1 = 9.1 L T1 = 273 + 42 = 315 K

P2 = 1.77 atm V2 = 11.5 Liter T2 = ?

T2 = 1.77 atm x 11.5 Liter x 315 K / 1.77 atm x 9.1 L

T2 = 398.07 K or 125.07 °C

The temperature of the gas sample at the new volume and pressure must be 125.07 °C.

Quest0ion D

PV= nRT

n = 0.514 atm x 2.27 L / 0.08206 L atm K-1 Mol-1 x 295 K

n = 0.048 Moles

Let us set P = 1 atm and T = 325 K

V = 0.048 x 0.08206 x 325 / 1 = 1.28 Liter

Hence the final gas volume will be smaller than 2.27 L.

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