Thermal & Kinetic Lecture 17

Back to the 1st law: PV cycles

LECTURE 17 OVERVIEW

Heat and work

PV cycles

Functions of state

Last time…

Entropy and the 2nd law

Reversible and quasistatic processes.

dS=dQ/T

Heat and work, engines and entropy

We have some big questions left to answer:

• How can we use thermal processes to do mechanical work?• What distinguishes heat from work?• What is the maximum efficiency we can expect

from a heat engine?• How is the maximum efficiency of a heat engine

related to entropy?

Work done by expanding gases: path dependence

Back in Section 2 of the module (Section 2.8 of the notes), we wrote down the following expression for the work done by an expanding gas:

dW = - PdV reversible process.

To get total work done we need to integrate: f

i

V

V

PdVW

dx

Work done by expanding gases: path dependence

T, V, and P are functions of state.Is W also a function of state?

V

Three different pathways: • Isothermal (constant temperature)• Isobaric (constant pressure)• Isochoric (constant volume)

P

Isotherm at T

1

2

V1 V2

P2

P1

Given that PV = nRT write down an expression for the work done by a gas when it expands from V1 to V2.

??ANS: W = - nRT ln(V2/V1)

!! Negative quantity because gas does work on surroundings.

f

i

V

V

PdVW

NB For reversible process.

Work done by expanding gases: path dependence

Isothermal expansion (1 2 on PV diagram) for ideal gas:

W = - nRT ln (V2/V1)

P

Isotherm at T

1

2

V1 V2

P2

P1

3

Write down an expression for the work done in an isobaric expansion of an ideal gas (3 2 on PV diagram).??ANS: W = -P2(V2 – V1)

Work done by expanding gases: path dependence

P

Isotherm at T

1

2

V1 V2

P2

P1

3

Isothermal expansion (1 2) for ideal gas:W = - nRT ln (V2/V1)Isobaric expansion (3 2) for ideal gas: W = -P2 (V2 – V1)

Write down an expression for the work done in an isochoric process involving an ideal gas (e.g. 1 3 on PV diagram).??

ANS: 0

Work done by expanding gases: path dependence

P

1

2

V1 V2

P2

P1

3

Shade in the region of the PV diagram that corresponds to the quantity of work done in an isobaric expansion from V1 to V2.

??

?? Now, shade in the region of the PV diagram that corresponds to the quantity of work done in an isothermal expansion from V1 to V2.

Unlike P, T, and V, work is not a function of state.

The amount of work depends on the path taken and there are an infinite number of paths connecting points 1 and 2 in the PV diagram above.

(…but there is an exception which we’ll come to soon).

Heat, work and the 1st law

The internal energy of an ideal gas is a function of temperature only. (Section 2a of the notes).

What is U for an isothermal compression of an ideal gas? What is the corresponding value for an isothermal expansion???

ANS: 0

Therefore, from the first law, how are the values of Q and W related in an isothermal process involving an ideal gas? ??

ANS: Q= -W

Consider a process involving isochoric heat transfer to 1 mole of an ideal gas.

Q = CV T = (3R/2) T

Gas

Q

Heat, work and the 1st law: isochoric heat transfer

T

Piston locked in place so V is constant.

If Q is (3R/2)T for the isochoric process shown above, by how much does the internal energy of the gas change? ??ANS: Isochoric, so W = 0, hence U is also (3R/2)T

Heat, work and the 1st law: Isobaric heat transfer

Gas

Q

T

Piston free to move so P is constant as gas expands from V1 to V2

?? Write down an expression for Q for this isobaric process.

ANS: Q = CPT = (5R/2) T

?? Write down an expression for the work done by the gas.

ANS: W = -P(V2 – V1)

Heat, work and the 1st law: Adiabatic compression of an ideal gas

In what we’ve considered thus far, heat was transferred into or out of the gas. What happens if compression of a gas is carried out under adiabatic conditions?

Gas

QXANS: U = W (because Q = 0)

For an adiabatic process, how are the change in internal energy and the work done on the gas related???

?? Is it possible for the temperature of the gas to rise if there’s no heat input?

ANS: dU = CVdT

If the temperature of an ideal gas changes by dT, by how much does its internal energy change???

Heat, work and the 1st law: Adiabatic compression of an ideal gas

Gas

If we can assume that the process is quasistatic and reversible, write down an expression for the amount of work done, dW, at each quasistatic step.

??ANS: dW = -PdV

So,CVdT = -PdV

However, we’re dealing with an ideal gas PV = RT (for 1 mole)

0

V

dVRTdTC

dVV

RTdTC

V

V

Heat, work and the 1st law: Adiabatic compression of an ideal gas

0

0

V

dVR

T

dTC

V

dVRTdTC

V

V

Integrate the above expression.??1lnln cVRTCV

Constant of integration.

We now divide through by CV to get: 2lnln cVC

RT

V

Heat, work and the 1st law: Adiabatic compression and expansion of an ideal gas

2lnln cVC

RT

V

However, we’ve shown in Section 2 that CP = CV + R (Eqn. 2.45). Hence, we can write the equation above as:

2lnln cVC

CCT

V

VP2ln)1(ln cVT

Rewriting this equation taking into account the properties of logs:

constTV 1

Using PV = RT again: constPV

Equation of an

adiabatic

PV curve for an adiabatic and an isothermal process

Adiabatic

Isotherm

At each point (P, V), the adiabatic for an ideal gas has a slope times that of an isotherm for an ideal gas.

P

V

Adiabatic work

• At the beginning of the 19th century it was assumed that heat was a substance called caloric which flowed between bodies.

• Prompted by measurements carried out by Benjamin Thompson, Joule wanted to determine the precise ‘form’ of heat.

http://lectureonline.cl.msu.edu/~mmp/kap11/cd295.htm

• Water stirred by falling weights turning paddle wheel.

• Water isolated from surroundings by adiabatic walls of container.

Adiabatic work

No matter how the adiabatic work was performed, it always took the same amount of work to take the water between the same two equilibrium states (whose temperatures differed by T)

If a thermally isolated system is brought from one equilibrium state to another, the work necessary to achieve this change is independent of the process used. (1st law, conservation of energy).

This seems to contradict what was said earlier –it appears that the work done in this case is path independent?

Adiabatic work is a function of state – the adiabatic work done is

independent of the path

U = Q + W. If Q=0 U = W

Adiabatic work

Work done by a gas in expanding adiabatically from a state (P1, V1) to a state (P2, V2)

11122

VPVPW

cPV

VPVP

2211

(c is a constant)