1. 1st Download the word document - guided notes thermodynamics from my website

2. Thermodynamics Chapter 10 pg 332 - 352

3. 1)In an engine How does the piston move?
Q  heat input
--More KE in gas, indicated by increase in temperature
-Thus more pressure
-Gas expands
-Volume increases
-Increase in volume = work done by system

4. Work = Force x change in distance
W = F Δ d ( d distance or h  height)
P = F/A so W = P A Δd
Volume : ΔV = A Δ d
W = PΔV
Work = Pressure x change in volume
Area for a circle  A= π r2
Volume for a cylinder  (V=πr2 h)

5. W = PΔV Work W = F Δ d W = P A Δ d
Assume that Pressure (P) is constant

6. Think 3) If gas expands : will change in volume (ΔV) be
( + )positive or ( - ) negative … explain
ΔV is (+)positive … ΔV = A Δ d = (df – di )  df > dv
4) If gas expands : will work (W ) done by system be
W done by system is ( + )positive…
W= PΔV because ΔV is positive

7. Isolated system Open system
A system is a portion of the universe that has been chosen for studying the changes that take place within it in response to varying conditions. A system may be complex, such as a planet, or relatively simple, as the liquid within a glass
Open system
Isolated system

8. Work done BY the system is (+)positive gas expands, thus ΔV is (+)positive
ΔV = A d

9. 5) Why is Work done ON the system negative ?

10. Why is Work done ON the system negative
Why is Work done ON the system negative ? Work done On the system is (+) Negative gas is compressed , thus ΔV is (- ) Negative
ΔV = A d

11. -- no displacement and no work is done
6) What is the work done when volume remains constant (no change in volume) ? Support your answer
Work is done only if the volume changes
Volume is constant
-- no displacement and no work is done
W = PΔV
ΔV = 0 Work = zero

12. An engine cylinder has a cross-sectional area of 0
An engine cylinder has a cross-sectional area of m2 How much work can be done by a gas in cylinder if the gas exerts a constant pressure of 7.5 x 105 Pa on the piston and moves the piston a distance of m?
Formula:
W = PΔV = P A d
Given :
A = .010 m2
P = 7.5 x 105 Pa
d = .04 m
Unknown
W=?
Plug and chug :
W = (7.5 x 105 Pa)(.010 m2)(.04 m )
W=300 J
W= 3.0 x102 J

13. The First Law of Thermodynamics Conservation of Energy

14. ΔU = Q – W
U Internal energy—the sum of the kinetic and potential energies of a system’s atoms and molecules. Can be divided into many subcategories, such as thermal and chemical energy. Depends only on the state of a system (such as its P , V, and T), not on how the energy entered the system
Q Heat —energy transferred because of a temperature difference. Characterized by random molecular motion.
W Work —energy transferred by a force moving through a distance. An organized, orderly process. Path dependent. W done by a system (either against an external force or to increase the volume of the system) is positive.

15. Internal Energy (U) decreases when
U, Q , W
Pg 339
Q positive +
System gains Heat
Q negative -
System loses heat
W positive +
Work done by system
expansion of gas
W negative -
Work done on system
compression of gas
Internal Energy (U) decreases when
Work done by a system
( expansion of gas)
Internal Energy (U) increases when
Work done on a system
(compression of gas)

16. Pg 339

18. 7) A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increased by 114 J during the process
A) What is the total amount of energy transferred as heat?
B ) Has energy been added to or removed from the refrigerant as heat?

19. Substituted the values in the equation Q = 114 J + (135 J) = -21 J
A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increased by 114 J during the process A) What is the total amount of energy transferred as heat
Given:
W = J
ΔU = 114 J
Unknown
Q = ?
Formula:
ΔU = Q – W
Rearrange the equation
Q = ΔU + W
Plug and chug :
Substituted the values in the equation
Q = 114 J + (135 J) = -21 J
Q = -21 J
Q is negative indicating that energy was removed from the system