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
Answer:
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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.