Question

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?

Answer 1

GIVEN :-

Initially ,

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

Finally ,

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

TO FIND :-

• Final Pressure (P₂)

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.

MORE TO KNOW :-

★ Boyle's Law :-

• P₁V₁ = P₂V₂

★ Charle's Law :-

• V₁/T₁ = V₂/T₂