Ideal Gases: dU=dW+dQ and dU=3/2RdT

[georg gill]

(I):

dU=dW+dQ

also (II):

$$dU=\frac{3}{2}RdT$$

if you compress a gas dW in dU is positive from pV=nRT lesser volume could either mean more pressure or more T. If dV gives dp only then dT=0 how can then dU for dW in (I) be equal dU in (II)?

 

[MrSid]

George; don't confuse your general equation of state (I) with the solved equation of state with respect to the setting of a constant T, p, or V or adiabatic condition (II).

What is implied is the question of what term in the general equation goes to zero when the solved equation is presented for a process; then using the conditions that create that solution, allows one to use the ideal gas law to show relationships of the variables that are not being held constant.

Do you see where this is going now?

 

[georg gill]

George; don't confuse your general equation of state (I) with the solved equation of state with respect to the setting of a constant T, p, or V or adiabatic condition (II).

What is implied is the question of what term in the general equation goes to zero when the solved equation is presented for a process; then using the conditions that create that solution, allows one to use the ideal gas law to show relationships of the variables that are not being held constant.

Do you see where this is going now?

I am a bit confused unfortunately. Why is (II) adiabatic?

thank you

 

[MrSid]

sorry George- not to imply that (II) is adiabatic;

but just to set the stage that one of those four conditions as constant when the equation of state is solved- In fact you will find in the wiki article Table_of_thermodynamic_equations
about halfway down an Equation Table for ideal gas and for the thermodynamic equation that matches (II) the variable that is kept constant is volume (isochoric).