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A meteorological balloon contains 250.0 l he at 22°c and 740.0mmhg if the volume of the balloon can vary according to external conditions what volume would it occupy at an altitude at which the temperature is -52°c and the pressure is 0.750 atm

Answer :

oladayoademosu

Answer:

243.15 l

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁ where P₁, V₁ and T₁ are the initial pressure, volume and temperature of the meteorological balloon and  P₂, V₂ and T₂ are the initial pressure, volume and temperature of the meteorological balloon.

P₁ = 740 mmHg V₁ = 250 l and T₁ = 22°C + 273 = 295 K,

P₂ =  0.750 atm = 0.750 × 760 mmHg = 570 mmHg (since 1 atm = 760 mmHg) and T₂ = -52°C + 273 = 221 K  

V₂ = P₁V₁T₂/P₂T₁ = 740 mmHg × 250 l × 221 K/570 mmHg × 295 K = 243.15 l

V₂ = 243.15 l

So the volume it would occupy at an altitude of temperature -52 °C where it pressure is 0.750 atm  is 243.15 l

The volume that should be occupied at an altitude is 243.15 l.

Gas law:

here the following equation should be used.

P₁V₁/T₁ = P₂V₂/T₂

So,

V₂ = P₁V₁T₂/P₂T₁

Here

P₁, V₁ and T₁ are the initial pressure, volume and temperature

and  P₂, V₂ and T₂ are the final pressure, volume and temperature

Now

P₁ = 740 mmHg

V₁ = 250 l

and T₁ = 22°C + 273 = 295 K,

And,

P₂ =  0.750 atm = 0.750 × 760 mm

Hg = 570 mmHg (since 1 atm = 760 mmHg)

and T₂ = -52°C + 273 = 221 K  

So,

V₂ = P₁V₁T₂/P₂T₁

= 740 mmHg × 250 l × 221 K/570 mmHg × 295 K

= 243.15 l

V₂ = 243.15 l

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