Answer to Question #174920 in Molecular Physics | Thermodynamics for Bahaa

Question

Find the change in the internal energy of one mole of a

monatomic ideal gas in an isobaric expansion at I atm from a volume

of 5m3 to a volume of 10m3. y for a monatomic ideal gas is 5/3

Answer 1

Let us write the ideal gas law:

$pV = \nu RT$ , where p is the pressure, V is the volume, $\nu$ is the amount of gas, R is the gas constant and T is the temperature.

For the initial and final conditions we get

$pV_1 = \nu R T_1, \;\; pV_2 = \nu RT_2.$

We may determine the initial temperature:

$1.01\cdot10^5\,\mathrm{Pa}\cdot 5\,\mathrm{m^3} = 1\,\text{mole}\cdot 8.31\,\mathrm{J/moles/K}\cdot T_1, \;\; T_1 = 60770\,\mathrm{K}$

It is enormously large value, because the amount of gas is very small for such a pressure. But we may assume that the gas in question is a part of stellar atmosphere in a very hot star (spectral class O)

For the final conditions $pV_2 = \nu R T_2, \; 1.01\cdot10^5\,\mathrm{Pa}\cdot 10\,\mathrm{m^3} =1\,\text{mole}\cdot8.31\,\mathrm{J/moles/K}\cdot T_2, \;\; T_2 = 121540\,\mathrm{K}.$

The change of internal energy is $\Delta U = \dfrac32 \nu R (T_2-T_1) = \dfrac32\cdot1\cdot8.31\cdot(121540-60770) = 7.6\cdot10^5\,\mathrm{J}.$

We may obtain this value without using the number of moles:

$\Delta U = \dfrac32 \nu R (T_2-T_1) = \dfrac32 \nu R \cdot \left(\dfrac{pV_2}{\nu R} - \dfrac{pV_1}{\nu R} \right) = \dfrac32(pV_2- pV_1) = \dfrac32p(V_2-V_1) = \dfrac32\cdot 1.01\cdot10^5\,\mathrm{Pa}\cdot (10\,\mathrm{m^3}-5\,\mathrm{m^3}) = 7.6\cdot10^5\,\mathrm{K}.$