Respuesta :
Answer:
The fraction of energy used to  increase the internal energy of the gas is 0.715
Explanation:
Step 1: Data given
Cv for nitrogen gas = 20.8 J/K*mol
Cp for nitrogen gas = 29.1 J/K*mol
Step 2:
At a constant volume, all the  heat will increase the internal energy of the gas.
At constant pressure, the gas expands and does work., if the volume changes.
Cp= Cv + R
⇒The value needed to change the internal energy is shown by Cv
⇒The work is given by Cp
To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp
Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol
Cv/Cp = 0.715
The fraction of energy used to  increase the internal energy of the gas is 0.715
Answer:
0.715 or 71.5%
Explanation:
The energy needed to change the temperature for an  isochoric process is equal to the change in internal energy.
Therefore;
[tex]q_v = \delta E\\\\q_v = nC_v \delta T[/tex]
Also; the energy needed to change the temperature for an isobaric process is equal to the change in enthalpy.
Therefore;
[tex]q__P}}= \delta H\\ \\q__P}} = n C_p \delta T[/tex]
However , the  fraction of the energy is used to increase the internal energy of the gas is calculated by the division of the total amount of energy needed for the isobaric process.
Thus;
[tex]\frac{\delta E}{\delta H}= \frac{nC_v \delta T}{nC_p \delta T}[/tex]
[tex]\frac{\delta E}{\delta H}= \frac{C_v }{C_p }[/tex]
From above; replacing the following value
[tex]C_v = 20. 8 \ J \ K^{-1} mol^{-1}\\\\C_p = 29.1 \ J \ K^{-1} mol^{-1}[/tex]
we have:
[tex]\frac{\delta E}{\delta H}= \frac{20.8 J \ K^{-1}mol^{-1} }{29.1 J \ K^{-1}mol^{-1} }[/tex]
= 0.715 or 71.5%
