On top of Mount Everest, the temperature is -19 ∘C in July. Being a physicist, you determine by how many degrees Celsius one needs to change the air temperature to double the average kinetic energy of its molecules. Part A Being a physicist, you determine by how many degrees Celsius one needs to change the air temperature to double the average kinetic energy of its molecules. ΔT = nothing ∘C

Let K be its average kinetic energy at t = -19°C = 273 + (-19) = 273 - 19 = 254 K = T. K = 3RT/2N = 3 × 8.314 J/mol-K × 254 K/(2 × 6.022 × 10²³ atoms/mol) = 5.26 × 10⁻²¹ J

When its average kinetic energy doubles, it becomes K₁ = 2K = 2 × 5.26 × 10⁻²¹ = 10.52 × 10⁻²¹ J at temperature T₂. So,

K₁ = 3RT₁/2N

T₁ = 2NK₁/3R

T₁ = 2 × 6.022 × 10²³ atoms/mol × 10.52 × 10⁻²¹ J/3 × 8.314 J/mol-K = 508 K

The temperature difference is thus ΔT = T₁ - T = 508 K - 254 K = 254 K.

Since temperature change in kelvin scale equals temperature change in Celsius scale ΔT = 254 °C

So, we need to change the temperature of the air by 254 °C to double its average kinetic energy.

hamzaahmeds hamzaahmeds · Answer 2 · 2021-11-05T10:26:52+01:00

This question involves the concepts of average kinetic energy, Boltzmann constant, and temperature.

The air temperature needs to be changed by "254°C" to double the kinetic energy.

The average kinetic energy of gas molecules is given by the following formula:

[tex]K.E = \frac{3}{2}KT[/tex]

where,

K.E = Average Kinetic Energy

K = Boltzmann Constant = 1.38 x 10⁻²³ J/K

T= Absolute temperature of gas

Now, for the initial condition:

[tex]K.E_1 = \frac{3}{2}(1.38\ x\ 10^{-23}\ J/K)(-19^oC+273)\\\\K.E_1 = 525.78\ x\ 10^{-23}\ J[/tex]

Now, for the final state the average kinetic energy is doubled:

[tex]K.E_2=2\ K.E_1 = \frac{3}{2}(1.38\ x\ 10^{-23}\ J/K)T_2\\\\T_2 = \frac{(2)(525.78\ x\ 10^{-23}\ J)}{2.07\ x\ 10^{-23}\ J/K}[/tex]

T² = 508 K = 235 °C

Hence, the change in temperature will be:

ΔT = T₁ - T₂ = 235°C + 19°C

ΔT = 254°C

Learn more average kinetic energy here:

https://brainly.com/question/1599923?referrer=searchResults

The attached picture shows the formula of average kinetic energy.