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Reil [10]
3 years ago
14

A balloon of helium gas in Alabama has a volume of 22.4 L at 26°C and a 1.03 ATM how big with a balloon get if is transported to

Denver on a sunny day when it is 22°C and has an air pressure of 0.81 ATM
Chemistry
1 answer:
Dimas [21]3 years ago
8 0

Answer:

The final volume of the balloon is = 28.11 L

Explanation:

Initial pressure P_{1} = 1.03 atm = 104.325 K pa

Initial temperature T_{1} = 26 °c = 299 K

Initial volume V_{1} = 22.4 L

Final temperature T_{2} = 22 °c = 295 K  

Final pressure P_{2} = 0.81 atm = 82 K pa

We know that

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Put all the values in above formula we get

\frac{(104.325)(22.4)}{299} = \frac{(82)(V_{2} )}{295}

V_{2} = 28.11 L

This is the final volume of the balloon.

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A balloon is filled with 12 L of air at a pressure of 2 atm. What is the volume of the balloon if the pressure is changed to 3 a
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Answer:

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Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant

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Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

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T = 22°C.

When the metal sample and water sample are mixed,

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Since, heat lost by metal is equal to the heat gained by water,

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Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

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Let Cp be the specific heat of the metal.

Substituting values,

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By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
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3 0
3 years ago
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