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physics help please?

1. A man fills a balloon with .5 liters of air. He then puts the balloon into the oven and heats it to 204•C. The temperature outside the oven was 20•C.

- 1A - what is the new volume of the balloon?

- 1B - what gas law is used to solve this?

2. A hot air balloon descends from a pressure of 380 mm Hg to 760 mm Hg and at the high altitude has a volume of 1,000 cubic meters. The initial temperature is 50•C and drops to 25•C.

-2A - convert Celsius temperature to kelvin ( absolute temperature). what are the new temperatures?

- 2B - what would be the final hot air ballon volume?

- 2C - which has law should be used determine the unknown (the final volume)?

Answer choices

1A - A. 3L

B. .3L

C. 8L

D. .8L

1B. A. Boyle's law

B. Charles' law

C. combined gas law

2A. A. T1 = 323K, T2=298K

B. T1=298K, T2=323K

C. T1=200K, T2=75K

D. T1=82 degrees K, Tx=57K

2B. A. 461 m^3

B. 250 m^3

C. 541 m^3

D. 46.1 m^3

2C. A. P1 x V1 = P2

B. V1/T1 = V2/T2

C. P1/T1 = P2/T2

D. P1 x V1/T1 = P2 x V2/T2

1 Answer

Relevance
  • 6 years ago

    1A.

    (0.5 L) x (204 + 273) K / (20 + 273) K = 0.8 L

    So answer D.

    1B. B. Charles' law

    2A.

    50•C + 273 = 323 K

    25•C + 273 = 298 K

    So answer A.

    2B.

    (1000 m^3) x (298 K / 323 K) x (380 mm Hg / 760 mm Hg) = 461 m^3

    So answer A.

    2C. D. P1 x V1/T1 = P2 x V2/T2

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