Is the internal energy of an ideal gas always $\frac{3}{2}nRT$? Is the internal energy of an ideal gas always $\frac{3}{2}nRT\,?$ I saw in Wikipedia that it is $\alpha nRT\;.$ Is $\alpha$ always equal to $\frac{3}{2}$ for an ideal gas?
 A: The answer to your query is Equipartition of Energy.
Equipartition Theorem:

At temperature $T\,_,$ the average energy of any quadratic degree of freedom is $\frac{1}{2} kT\;.$

For each degree of freedom, the ideal gas molecule can store $\frac12 kT$ of energy on average.
For monatomic ideal gas molecule, there are only three degrees of freedom: translational kinetic energies for moving along three perpendicular axes. Therefore, the molar heat capacity at constant volume $C_V$ for monatomic gas is $$C_V= \frac32 R \;.$$
Similarly, for a diatomic molecule, there are five degrees of freedom: three translational kinetic energy; two rotational kinetic energies. It can rotate about two different axes (not along the internuclear axis). Therefore, $$C_V= \frac52 R\;.$$
Internal energy of an ideal gas is given by $$E_\textrm{int} = nC_VT\;.$$
For monatomic ideal gas, $$E_\textrm{int} = n \frac32 R T\,.$$
For diatomic ideal gas,  $$E_\textrm{int} = n \frac52 R T\,.$$
So, in general, $$E_\textrm{int} = n \alpha RT\,.$$
A: It depends on gas whether its monoatomic diatomic nonrigid ,rigid as $\alpha$ is degree off freedom which depends on gas for mono it s $3/2$
A: It depends on what one defines as an ideal gas.  In the physics literature, an ideal gas is defined as one which has constant heat capacity from absolute zero all the way up to the temperature T.  In the engineering literature, we consider an ideal gas to have temperature-dependent heat capacity, just as real gases do at low pressures.  In both cases, the heat capacity is considered to be independent of pressure.  Since this is a physics exchange, there may not be much interest in the engineering treatment, but it is still important to understand the distinction if one is consulting the literature.
