[Araconan]

When studying the thermodynamics behind gases, my textbook states that the Energy "heat" required to change the energy of 1 mole of an ideal gas by ΔT is 3/2 RΔT, which was derived from the equation (KE)_avg = 3/2 RT. So is it then correct to state that q = 3/2 RΔT per mole? (Where q can then be q_v or q_p depending on whether the gas is heated at a constant pressure or a constant volume)

Furthermore, after considering the two scenarios of heating the gas at a constant volume and a constant pressure, deriving the equations C_v  = 3/2R (per mole, increasing by 1K), and C_p = C_V + R = 5/2R (per mole, increasing by 1K), it goes on to state that for an ideal gas, E = 3/2 RT. This is the part that I don't understand. Previously, it was just stated that KE = 3/2 RT, and since E = KE + PE, how can E = 3/2 RT?

Also, it states that q_p = C_pΔT (per mole), which can then be written as q_p = C_v + RΔT, which is the same as saying q_p = ΔE + PΔV. How can this possibly be true? Isn't ΔE = q+w? Even if w = 0, q = ΔE. So how can q_p = ΔE + PΔV?

[Jorriss]

Furthermore, after considering the two scenarios of heating the gas at a constant volume and a constant pressure, deriving the equations C_v  = 3/2R (per mole, increasing by 1K), and C_p = C_V + R = 5/2R (per mole, increasing by 1K), it goes on to state that for an ideal gas, E = 3/2 RT. This is the part that I don't understand. Previously, it was just stated that KE = 3/2 RT, and since E = KE + PE, how can E = 3/2 RT?

An ideal gas has no potential energy. A monotomic ideal gas has only translational energy. The equation E=1.5RT is not exact and only considers translational energy.

Also, it states that q_p = C_pΔT (per mole), which can then be written as q_p = C_v + RΔT, which is the same as saying q_p = ΔE + PΔV. How can this possibly be true? Isn't ΔE = q+w? Even if w = 0, q = ΔE. So how can q_p = ΔE + PΔV?

q_p = ΔE + PΔV
Notice that at constant p PΔV=-w so that is still just the first law.