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 an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K10?

Answer 1

Answer:

1. $195.27\times10^{-23}J$

2. 566 K

Explanation:

1. Temperature of the ideal gas is 10°C. convert the temperature into Kelvins.

T = (10+273) = 283 K

Average kinetic energy for 1 degree of freedom is, $K.E. =\frac{1}{2}KT$

where, K is Boltzmann constant.

Substitute the values:

$K.E. =\frac{1}{2}KT=\frac{1}{2}(1.38\times 10^{-23} m^2 kg s^{-2} K^{-1}) (283K)=195.27\times10^{-23}J$

2.

Average kinetic energy is twice. $K.E = 2\times 195.27\times10^{-23}J=390.54\times10^{-23}J$

$T = \frac{2 (K.E)}{K}\\ T = \frac{2\times 390.54\times10^{-23}}{1.38\times 10^{-23}}\\ T = 566 K$

Answer 2

Answer:

          T ’’ = 293º C

Explanation:

Here we must equal the kinetic energy with the thermal energy of the Boltzmann equation

       E = K

       E = k T

Where k is the Boltzmann constant and T the absolute temperature

   

       K = k T

T ’= 10ºC, the equation to move to Kelvin degrees is

      T = (T ’+273) [K]

Let's use the first data

     K10 = k T1

We write the second data

    2 (K10) = k T2

 

Let's divide the two equations

    2 (K10) / (K10) = k T2 / k T1

    2 = T2 / T1

    T2 = 2 T1

Now let's write this value in degrees Celsius

     (T ’’ + 273) = 2 (T ’+ 273)

     T ’’ = 2T ’+ 2 273 -273

     T ’’ = 2 T ’+ 273

     T ’’ = 2 10 + 273

     T ’’ = 293º C

This is the temperature to double the kinetic energy