Physics, asked by cristyarangosmsosa, 1 month 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 Anonymous
1

GIVEN :-

Initially ,

  • Pressure (P₁) = 1000mmHg
  • Volume (V₁) = 30000 litres
  • Temperature (T₁) = 40°C

Finally ,

  • Volume (V₂) = 30000 + 2500 = 32500litres
  • Temperature (T₂) = 70°C

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TO FIND :-

  • Final Pressure (P₂)

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TO KNOW :-

By Combined Gas Law ,

 \\  \bigstar \boxed{ \sf \:  \dfrac{ P_{1} V_{1}}{T_{1}}  =  \dfrac{P_{2}V_{2}}{T_{2}} } \\  \\

SOLUTION :-

We have ,

  • P₁ = 1000mmHg
  • V₁ = 30000 litres
  • T₁ = 40°C
  • V₂ = 32500 litres
  • T₂ = 70°C

Putting values in formula ,

 \\  \implies  \sf \:  \dfrac{1000 \times 30000}{40}  =  \dfrac{P_{2} \times 32500}{70}  \\  \\  \\  \implies \sf \:  \dfrac{3000000 \cancel0}{4 \cancel0}  =  \dfrac{P_{2} \times 32500}{70}  \\  \\  \\ \implies  \sf \: P_{2} =  \dfrac{3000000 \times 70}{32500 \times 4}  \\  \\  \\  \implies \sf \: P_{2} =  \dfrac{2100000 \cancel{00}}{325 \cancel{00} \times 4}  \\  \\  \\   \implies\sf \: p_{2} =  \dfrac{2100000}{1300}  \\  \\  \\    \implies\boxed{\sf \: P_{2} =  1615.38mmHg} \\  \\

Hence , Pressure becomes 1615.38mmHg.

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MORE TO KNOW :-

Boyle's Law :-

  • P₁V₁ = P₂V₂

Charle's Law :-

  • V₁/T₁ = V₂/T₂
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