On expanding isothermally from 2L to 4L, an ideal gas does 6J of work, as the pressure drops from 2 atm to 1 atm. By how much must it expand to do an additional 6J of work? A] It must expand an additional 2L, to reach 6L B] It must double again, to 8L C] It must increase four-fold, to 16L D] It can do no more work, as it has reached 1 atm. E] Cannot determine without knowing T

An ADIABATIC PROCESS is one in which no heat transfer occurs. Q=0. If we expand a gas from V A to V B adiabatically, what will be the final pressure? A] More than P B B] Less than P B C] Equal to P B D] cannot determine Let’s do the math

Review questions: For the constant pressure ideal gas process shown, The change in internal energy U of the gas is A] nC v  B] nC p  C] 0 D] cannot determine

If the gas is monatomic and ideal, for the constant pressure process shown, the change in internal energy can also be expressed as: A] p  V B] C] D] (use a paper & pencil. Hint: what is  T?)

For the cyclic process shown, W is: A] 0, because it’s a loop B] p 0 V 0 C] - 2p 0 V 0 D] 2 p 0 V 0 E] 6 p 0 V 0 For the cyclic process shown,  U is: For the cyclic process shown, Q is: For ONE cycle: Monatomic, Ideal

A] 12 only B] 23 only C] 34 only D] 12 & 23 E] 12 & 34 Along which sides is heat added to the gas?

A] 34 only B] 41 only C] 34 & 41 D] heat is not removed on any side Along which sides is heat removed from the gas?

A] added B] removed C] the same amount of heat is removed as was added We know heat is added 12 and 23. Heat is removed on 34 and 41. Is more heat added, or removed?

A] added B] removed C] the same amount of heat is removed as was added We know heat is added 12 and 23. Heat is removed on 34 and 41. Is more heat added, or removed? Heat is NOT a state variable. It is a process variable. More heat was added than removed, because the gas did work and energy is conserved!

We saw that there is always heat “rejected” in a thermodynamic cycle = Q c