Question

Calculate the total energy of one mole of an ideal monatomic gas at $27^\circ C$.
A. $900cal$
B. $1800cal$
C. $7482cal$
D. $0cal$

Answer 1

Hint: The total energy of a given quantity of an ideal gas can be estimated using the kinetic theory of gases, from which we get a relationship relating the proportionality between the total energy of an ideal gas and the absolute temperature.

Complete step by step answer:
From the postulates of the kinetic theory, we have an expression for the pressure of an ideal gas as follows:
$P = \dfrac{1}{3}\dfrac{{Nm{{({u^2})}_{avg}}}}{V}$
$ \Rightarrow m{({u^2})_{avg}} = \dfrac{{3PV}}{N}$ (1)
Where $P$ is the pressure, $N$ is the number of particles, $m$ is the mass of the individual particle, ${({u^2})_{avg}}$ is the average velocity term of the gas particle and $V$ is the volume occupied.
We know that the total kinetic energy of a particle is written as $E = \dfrac{1}{2}m{({u^2})_{avg}}$ (2)
Substituting the value from equation (1) into equation (2), we get:
$E = \dfrac{1}{2} \times \dfrac{{3PV}}{N}$ (3)
But from our knowledge of the ideal gas theory, we know that $PV = nRT$
Where $n$ is the number of moles of the ideal gas, $R$ is the universal gas constant and $T$ is the absolute temperature. Also, in this case, as we are asked to find the energy of one mole of particles, $N$ is equal to one mole. Therefore, we can replace the $N$ term with one mole. Hence, by substituting these values into equation (3), we get:
$E = \dfrac{3}{2}\dfrac{{nRT}}{n}$ since $N = n = 1mol$
Simplifying the above equation, we get the equation for energy for one mole of an ideal gas:
$E = \dfrac{3}{2}RT$
Notice that in the options given, all values are given in units of $cal/mol$. Therefore, we must choose the appropriate value of $R$ (the one in which units are in terms of $cal/mol$). As we know, such a relation does exist: $R = 2cal/mol/K$.
So, for one mole, $R = 2cal/K$
We must also convert the temperature from degree Celsius to the Kelvin scale:
$T = 27 + 273 = 300K$
Substituting these values into the equation for total energy, we get:
$E = \dfrac{3}{2} \times 2\dfrac{{cal}}{K} \times 300K$
On simplifying, we get the final answer as:
$E = 900cal$

Thus, the correct option is option A.

Additional Information: It may be useful to go through the postulates of the kinetic theory for a concrete understanding of the interdependence between the gas constant, temperature and various properties of an ideal gas like root mean square velocity, pressure, velocity etc.

Note: The energy here is completely kinetic, as postulated from the kinetic theory, and is by virtue of the random motion of the gas particles. The equation for internal energy derived here is only applicable to ideal monatomic gases. For gases with more number of atoms, total energy keeps increasing linearly by a magnitude equal to the value of the gas constant.We can also use different units of $R$, like the SI unit and still get the answer. In this case, we must convert our final answer from Joules to calorie. The only thing to keep in mind is to maintain the consistency of units.