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Thermodynamics Chapter 10 pg 332 - 352

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

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)

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

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

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

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

5) Why is Work done ON the system negative ?

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

-- 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

An engine cylinder has a cross-sectional area of 0 An engine cylinder has a cross-sectional area of 0.010 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 0.040 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

The First Law of Thermodynamics Conservation of Energy

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.

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)

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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?

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 = - 135 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