Question

For nitrogen gas the values of Cv and Cp at 25°C are 20.8 J K−1 mol−1 and 29.1 J K−1 mol−1, respectively. When a sample of nitrogen is heated at constant pressure, what fraction of the energy is used to increase the internal energy of the gas?

Answer 1

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 2

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;

$q_v = \delta E\\\\q_v = nC_v \delta T$

Also; the energy needed to change the temperature for an isobaric process is equal to the change in enthalpy.

Therefore;

$q__P}}= \delta H\\ \\q__P}} = n C_p \delta T$

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;

$\frac{\delta E}{\delta H}= \frac{nC_v \delta T}{nC_p \delta T}$

$\frac{\delta E}{\delta H}= \frac{C_v }{C_p }$

From above; replacing the following value

$C_v = 20. 8 \ J \ K^{-1} mol^{-1}\\\\C_p = 29.1 \ J \ K^{-1} mol^{-1}$

we have:

$\frac{\delta E}{\delta H}= \frac{20.8 J \ K^{-1}mol^{-1} }{29.1 J \ K^{-1}mol^{-1} }$

= 0.715 or 71.5%