Question

When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are using. In general, you should use the Kelvin scale (for which T=0 represents absolute zero) in such calculations. This is because the standard thermodynamic equations (i.e., the ideal gas law and the formula for energy of a gas in terms of temperature) assume that zero degrees represents absolute zero. If you are given temperatures measured in units other than kelvins, convert them to kelvins before plugging them into these equations. (You may then want to convert back into the initial temperature unit to give your answer.)

Part A)

The average kinetic energy of the molecules of anideal gas at 10 degree C has the value K10. At what temperature T1 (in degrees Celsius) will the average kinetic energy ofthe same gas be twice this value, 2K10?
Express the temperature to thenearest integer.

Part B) The molecules in an ideal gas at 10 degree C have a root-mean-square (rms) speed Vrms. At what temperature T2 (in degrees Celsius) will the molecules have twice therms speed, 2Vrms ?
Express the temperature to thenearest integer.

Answer 1

A) T₁ = 293°C

B) T₂ = 858.8°C

Step-by-step explanation:

Part A)

The average kinetic energy of an ideal gas molecule is given by the formula:

Kinetic Energy = K.E = (3/2)KT

where,

K = Boltzman's Constant

T = Absolute Temperature

For initial state:

T = 10° C + 273 = 283 K

K.E = K₁₀

Therefore,

K₁₀ = (3/2)K(283)

K₁₀ = 424.5 K

Now, for final state,

K.E₁ = 2 K₁₀ = 2 (424.5 K) = 849 K

T₁ = ?

Therefore,

849 K = (3/2) KT₁

T₁ = (849*2)/3 Kelvin

T₁ = 566 Kelvin = 293° C

Part B)

The average kinetic energy of an ideal gas molecule is given by the formula:

Kinetic Energy = K.E = (3/2)KT

Also,

K.E = (1/2)mv²

Comparing both equations we get:

(1/2)mv² = (3/2)KT

v = √(3KT/m)

For initial state:

T = 10° C + 273 = 283 K

v = Vrms

Therefore,

Vrms = √(3K*283/m)

Vrms = 29.14 √(K/m)

Now, for final state,

v = 2 Vrms = 2 (29.14 √(K/m)) = 58.27 √(K/m)

T₂ = ?

Therefore,

58.27 √(K/m) = √(3KT₂/m)

T₂ = (58.27)²/3 kelvin

T₂ = 1131.8 Kelvin = 858.8° C