Solution to Assignment questions JIF 314 Thermodynamics

Based on the text book

Heat and thermodynamics by Zemansky and Dittman, 7th edition, Mcgraw-Hill. Chapter 1 Problem 1.1. Solve using Excel.

First, calculate the value θ of the gas: 273.1KTP

PP

θ⎛ ⎞

= ⎜ ⎟⎝ ⎠

.

PTP (kPa) P (kPa) θ (K) 33.331 51.19 419.521178566.661 102.37 419.486494499.992 153.54 419.4434195133.32 204.69 419.390342

theta versus P_{TP}

y = -0.0436x + 419.57

300320340360380400420440

33.331 66.661 99.992 133.32P_{TP}, kPa

thet

a (K

)

the line intercepts with the vertical axis at y = 419.57

θ vs. PTP is a straight line in the form of y = mx + c, where y ª θ , xª PTP. The value of θ when PTP becomes zero is the value of the temperature of the gas. This value is simply the value of intersection, c, in the formula of the straight line in the form of y = mx + c. From the formula of the straight line generated by Excell, the intersection of the straight line is c = 419.57 in the graph of TP . vs Pθ . Hence, the temperature of the gas in the bulb is θ = 419.57 K.

Problem 1.3. (a) The temperature with resistance measured to be 1000Ω can be calculated using the relationship between R′ and T, as per

log log , 1.16, 0.675R a b R a bT

′′= + = − = .

Setting R′ = 1000 Ω,

( )

( ) ( ) ( ) ( )( )

( ) ( ) [ ] ( )( )( )

2

2 2 2

2 2 2

22 2

22 2

log loglog (log ) 2 log

log(log ) 2 log

log 1000

1.16 0.675 log 1000 2 1.16 0.675 log

0.6908 1.441.16 0.675 0.6908 2 1.16 0.675 0.6908

R Ra b R a b R ab RT T

RTa b R ab R

R

⎛ ⎞′ ′′ ′ ′= + = = + +⎜ ⎟⎜ ⎟

⎝ ⎠′

=′ ′+ +

=′− + + −⎡ ⎤⎣ ⎦

= =− + + −

Hence, the temperature of the helium cryostat is 1.44 K. (b) Use Excell. Plot log R′ vs. log T graph by forming the following table:

R’ log R’ T = log R’/(a + b log R’)^2 log T

1000 6.907755 0.563018189-

0.57444

5000 8.517193 0.404427271-

0.90528

10000 9.21034 0.360158153-

1.02121

15000 9.615805 0.338393713-

1.08355

20000 9.903488 0.32444907-

1.12563

25000 10.12663 0.31438398-

1.1571430000 10.30895 0.306603264 -1.1822

Problem 1.9: ( ) ( ) ( )9 9F C 32 99.974 32 211.95 F5 5

θ θ= + = + = (5 significant figures).

log T vs log R'

-2

-1.5

-1

-0.5

06.0 7.0 8.0 9.0 10.0 11.0

Chapter 2 Problem 2.1

(a) Given the equation of state for a ideal gas PV = n RT, show that 1T

β = .

Solution: Given equation of state for a ideal gas

PV = n RT, Eq. (1)

and the definition of volume expansivity 1 VV T

β ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠, it is easily verified that β = 1/T by taking the partial

derivate of Eq. (1) with respect to T:

( ) VPV nRT P nRT T∂ ∂

= → =∂ ∂

Eq. (2)

Inserting PV = nRT into Eq. (2), we arrive at

1V nR PV VT P T P T

∂= = =

∂

Hence, 1 VV T

β ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠= 1 1V

V T Tβ ⎛ ⎞= =⎜ ⎟

⎝ ⎠.

(b) Show that the isothermal compressivility κ = 1/P. Solution Given equation of state for a ideal gas

PV = n RT, Eq. (1)

and the definition of isothermal compressibility 1 1 VB V P

κ ∂⎛ ⎞= = − ⎜ ⎟∂⎝ ⎠, it is easily verified that β = 1/P by

taking the partial derivate of Eq. (1) with respect to P:

( ) ( ) 0VPV nRT P V nRTP P P∂ ∂ ∂

= → + = =∂ ∂ ∂

Eq. (2)

Inserting PV = nRT into Eq. (2), we arrive at

V VP P

∂= −

∂

Hence, 1 1 1V VV P V P P

κ ∂⎛ ⎞ ⎛ ⎞= − = − − =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠.

Problem 2.2: Given the equation of state of a van der Waals gas, ( )2

aP v b RTv

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

, calculate

(a) T

Pv

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

, (b) v

PT

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

.

Solution: (a) Taking the partial derivative with respect to v, with constant T,

( ) ( )2 0TT

aP v b RTv v v

∂ ⎡ ⎤ ∂⎛ ⎞+ − = =⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠⎣ ⎦

( ) ( )

( )

2 2

3 2

2

3

0

2 0

2

TT

T

T

a av b P P v bv v v v

P a av b Pv v v

aPP avv v b v

∂ ∂⎛ ⎞ ⎛ ⎞− + + + − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂ ⎛ ⎞− − + + =⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠⎝ ⎠

+∂= − +

∂ −

(b) Taking the partial derivative with respect to T, with constant v,

( ) ( )

( ) ( )

( )

( )

2

2 2

2 2

2

1

0 0

vv

vv

v vv

v

v

aP v b RTT v T

a av b P P v b RT v v T

P a vv b a P RT T v v T

P av b P RT v

P RT v b

∂ ⎡ ⎤ ∂⎛ ⎞+ − =⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠⎣ ⎦

∂ ∂⎛ ⎞ ⎛ ⎞− + + + − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞− + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞∂ ⎛ ⎞− + + + ⋅ =⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠⎝ ⎠

∂=

∂ −

(c)

2 3 2

3

1212

T P v

V

P

T

P v Pv T T

P Rv RT v b

a ab aPT PP av v P v Pvv b v

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟∂ ∂⎛ ⎞ ⎝ ⎠ −→ = = − = ⎜ ⎟⎜ ⎟ ∂∂ ⎛ ⎞⎝ ⎠ ⎜ ⎟+ + −⎜ ⎟ ⎝ ⎠∂⎝ ⎠ − +

−

Problem 3.2 (a) Show that the work done by an ideal gas during the quasi-static, isothermal expansion from an initial pressure Pi to a final pressure Pf, is given by W = nRT ln (Pf /Pi). Solution: For isothermal process, PiVi = PfVf. Hence Vi /Vf = Pf /Pi. Substitute this into W = -nRT ln (Vf /Vi ), we get W = -nRT ln (Pi /Pf )= nRT ln (Pf /Pi). Problem 3.3 An adiabatic chamber with rigid walls consists of two compartments, one containing a gas and the other evacuated; the partition between the two compartments is suddenly removed. Is the work done during an infinitesimal portion of this process (called an adiabatic expansion) equal PdV ? Answer: NO. Because there is no work done against the expansion of the gas-filled compartment by the evacuated compartment.