Physics, asked by cristyarangosmsosa, 2 months ago

A hot air balloon has a volume of 30,000 liters of air at a temperature of 40,000° C and a pressure of 1,000 mm Hg. If the temperature in the balloon is increased to 70° C while the volume increases by 2500 liters, what is the resulting pressure?

Answers

Answered by abhishekkhillare1010
1

Answer:

*(1,0,0,0,3) ही सहगुणक रूपातील बहुपदी 'x' हे चल वापरून घात रुपात पुढीलप्रमाणे लिहितात.*

1️⃣ x² + 3

2️⃣ x⁴ +3

3️⃣ x + 3

4️⃣ x-3

Answered by dualadmire
1

The correct question is:

A hot air balloon has a volume of 30,000 liters of air at a temperature of 40,000° C and a pressure of 1,000 mm Hg. If the temperature in the balloon is increased by 70° C while the volume increases by 2500 liters, what is the resulting pressure?

Given:

Volume of the hot air balloon = 30,000 liters

Temperature of the hot air balloon= 40,000°C

Pressure of the hot air balloon= 1000 mm Hg

Increase in temperature = 70°C

Increase in volume = 2500 liters

To find:

The resulting pressure.

Solution:

The final temperature = 40,000 + 70= 40,070

The final volume = 30,000 + 2500 = 32,500

Using the combined equation of Boyle's law and Charle's law:

P1V1 / T1 = P2V2/ T2

P2 = P1V1T2 / T1V2

= 1000×30,000×40,070 / 40,000×32500

= 924.69 mm Hg

Therefore the resultant pressure is 924.69 mm Hg.

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