A helium balloon in a closed car occupies a volume of 1.32 L at 30.0°C. If the car is parked on a hot day and the volume of the balloon increases to 2.50 L what is the temperature in the car, assuming the pressure remains constant?

T₂ = 573.86 K

Explanation:

Given data:

Initial volume = 1.32 L

Initial temperature = 30.0°C (30+273 = 303 K)

Final volume = 2.50 L

Final temperature in car = ?

Solution:

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.50 L × 303 K / 1.32 L

T₂ = 757.5 L.K / 1.32 L

T₂ = 573.86 K

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

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