Question

If the temperature outside is 60°F, what is the approximate temperature on the Celsius scale and the Relvin scale?

If the air temperature drops to 30°F during the night, how has the kinetic energy of the air particles changed?

Answer 1

KE at 30°F=0.942 KE at 60°F

Further explanation

The average kinetic energy value is only affected by temperature changes. The higher the temperature, the average kinetic energy of the molecule increases

This molecule is very small when compared to the distance between molecules, so the volume of gas contains mostly empty space

Gas particles move randomly (both speed and direction , as vector)

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

`\large{\boxed{\bold{v_(rms)=\sqrt{(3RT)/(Mm) } }}`

R = gas constant, T = temperature, Mm = molar mass of the gas particles

Kinetic energy :

`\tt KE=(1)/(2)mv^2`

The temperature outside is 60°F

Convert to Celcius and Kelvin

°C = 5/9 (°F-32)

`\tt ^OC=(5)/(9)(60-32)=15.56`

°K = °C + 273

`\tt ^oK=15.56+273=288.56`

For 30°F :

`\tt ^oC=(5)/(9)(30-32)=-1.1\\\\^oK=-1.1+273=271.9`

• KE at 288.56 °K(60 °F)

`\tt KE_1=(1)/(2)m.(3R.288.56)/(Mm)`

• KE at 271.9°K(30°F)

`\tt KE_2=(1)/(2)m.(3R.271.9)/(Mm)`

`\tt (KE_2)/(KE_1)=(271.9)/(288.56)=0.942\\\\KE_2=0.942KE_1`