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jeyben [28]
2 years ago
13

A balloon is inflated to a volume of 7.0 L 7.0 L on a day when the atmospheric pressure is 765 mmHg. The next day, as a storm fr

ont arrives, the atmospheric pressure drops to 737 mmHg 737 mmHg . Assuming the temperature remains constant, what is the new volume of the balloon, in liters?
Chemistry
1 answer:
kirill115 [55]2 years ago
6 0

Answer:

The new volume is 7.27 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 7.0 L

The pressure = 765 mmHg = 765/760 atm = 1.00658 atm

The pressure drops to 737 mmHg = 737/ 760 atm = 0.969737

The temperature remains constant

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure in the balloon = 765 mmHg

⇒with V1 = the initial volume = 7.0L

⇒with P2 = the decreased pressure = 737 mmHg

⇒with V2 = the new volume = TO BE DETERMINED

765 mmHg * 7.0 L = 737 mmHG * V2

V2 = (765 mmHG * 7.0 L) / 737 mmHG

V2 = 7.27 L

The new volume is 7.27 L

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Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

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Step 2: Calculate temperature

V1 / T1 = V2 /T2

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This question is incomplete because the options are missing; here are the options:

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