Question

The densities of air at −85 °C, 0 °C, and 100 °C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data, and assuming that air obeys Charles’ law, determine a value for the absolute zero of temperature in degrees Celsius.

Answer 1

Answer:

-271,4°C

Explanation:

Charles's law says that when pressure of a gas held constant, temperature in kelvin and volume will be in a direct proportion. That is:

V/T=k

As density = mass/volume:

density⁻¹/T = k*mass -Where k*mass is a constant-

The temperature in kelvin must be (X+T) where X is the value of °C in the absolute zero scale.

For the three values:

1,877⁻¹/(X-85°C) = 1,294⁻¹/(X) = 0,946⁻¹/(X+100°C)

That is:

(1) 1,877⁻¹/(X-85°C) = 1,294⁻¹/(X)

(2) 1,294⁻¹/(X) = 0,946⁻¹/(X+100°C)

(3) 1,877⁻¹/(X-85°C) = 0,946⁻¹/(X+100°C)

Solving X for (1):

1,877⁻¹X= 1,294⁻¹X - 65,7°C

-0,24X = - 65,7°C

X = 269,5°C

For (2): X = 271,8°C

For (3): X = 273,0°C

The average of these values is:

269,5°C+271,8°C+273,0°C / 3= 271,4

As 271,4 is the value of 0°c, the absolute zero is -271,4

I hope it helps!