Respuesta :
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
Answer:
1. [tex]195.27\times10^{-23}J[/tex]
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, [tex]K.E. =\frac{1}{2}KT[/tex]
where, K is Boltzmann constant.
Substitute the values:
[tex]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 [/tex]
2.
Average kinetic energy is twice. [tex] K.E = 2\times 195.27\times10^{-23}J=390.54\times10^{-23}J[/tex]
[tex] T = \frac{2 (K.E)}{K}\\ T = \frac{2\times 390.54\times10^{-23}}{1.38\times 10^{-23}}\\ T = 566 K[/tex]
