Question

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. 20) Assuming that you can treat the gas as ideal gas and that the volume of the liquid compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas) approximately zero,
a. 9.4 x 10-21 J
b. 3.8 х 1027 J
c. 4.2 x 10-18 J
d. 10-20 J e. 2.1 x 10-19 J

Answer 1

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

$Q=\Delta U+W$

$Q=\Delta U+nRT$

Put the value into the formula

$4480=\Delta U+\dfrac{20}{28}\times8.314\times77$

$\Delta U=4480-\dfrac{20}{28}\times8.314\times77$

$\Delta U=4022.73\ J$

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

$N=n\times \text{Avogadro number}$

Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$