Question

4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-mmHg and its size adjusts to 14.3-L. What Celsius temperature was the wind? ​

Answer 1

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

What is Combined gas law?

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

• Initial volume V₁ = 14.5L

• Initial pressure P₁ = 0.980atm

• Initial temperature T₁ = 20.0°C = 293.15K

• Final volume V₂ = 14.3L

• Final pressure P₂ = 740.mmHg = 0.973684atm

• Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

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