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100 POINTS! Final Honor Activity Question
An unbreakable meteorological balloon is released from the ground. Ground level pressure is 98.5 kPa and the temperature is 18oC. The balloon contains 74.0 dm3 of hydrogen gas. As the balloon ascends, the pressure drops to 7.0 kPa.
What is the new volume of the balloon, assuming no temperature change?

If a temperature drop of 80oC occurs, what is the new volume of the balloon?

Of the two factors, pressure and temperature, which had the greatest effect in changing the volume of the balloon? Support your answer with your calculations.

Approximately 160 km above the Earth the pressure of the atmosphere is about 3.00 x 10-4 Pa and the temperature is about -180oC. If 100 cm of a gas at STP is released at this distance from the Earth, what volume would it occupy?

Sagot :

pstnonsonjoku

The change in temperature had the greatest effect at changing the volume of the balloon.

What are the gas laws?

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 =  74.0 dm3

P2 =  7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 =  98.5 kPa *  74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

T1 = 291 K

V2 = ?

T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 =  100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 =  3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 =  1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:https://brainly.com/question/12669509

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