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3 years ago

6

a hot air balloon has a volume of 875L. what is the original temperature of the balloon if its volume changes to 955 L when heat

ed to 56

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

using Charles law

v1/t1= v2/ t2

Explanation:

875/t1=955/56+273

t1= 329* 875/955

t1= 302.82k

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Likurg_2 [28]

Answer:

1.389

Explanation:

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3 years ago

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GuDViN [60]

Answer:

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Explanation:

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We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.

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vova2212 [387]

Answer:

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