1

Chem340 Physical Chemistry for Biochemists Exam 1 Feb 11, 2011 Your Name _ I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class. Signature . Sec 1 Multiple Choice [ /42] Sec 2 Calculation/MC [ /24] Sec 3 Long Question [ /40] Sec 4 Derivation [ /10] Total [ /100] Total points are 124 points. If you score more than 100, you will receive 100 points.

2

Multiple Choice Questions (Some questions may be similar to those one in quiz, but not the same) Each 3 points [ /42] Q1. Which of the following P-V cycles has the largest work in absolute value (namely, |w|)? Assume that Pext = Pint. Choose all the correct answers. (a) [P (bar),V (L) ] [1, 1] [1, 3] [3, 3] [3, 1] [1, 1] (b) [P (bar),V (L) ] [1, 1] [3, 1] [3, 3] [1, 3] [1, 1] (c) Isothermal compression from [P (bar),V (L)] of [3, 1] to [1, 3] and then isothermal expression from [1, 3] to [3, 1] (d) Isothermal compression from [P (bar),V (L)] of [1, 3] to [3, 1]. Then, [P,V] =[3, 1] [3, 3] [1, 3] Q2. Select two correct statements from the following choices. (a) “Equilibrium” means the situation in which a system parameter reaches constant over time in any part of the systems (b) “Equilibrium” means the situation in which a system parameter reaches constant over time in some parts of the systems (c) The zeroth law of thermodynamics states that when two systems are separately in thermal equilibrium with the third system, these two systems are also in thermal equilibrium (d) The first law of thermodynamics is equivalent to the fact that no heat is generated in the reversible process. (e) In a reversible process, Pext = -Pint Q3. Calculate the average translational energy <mv2/2> for one molecule of H2O assuming that it is an ideal gas, where m is a mass of the molecule and v is a velocity. Choose the closet from the choices (a- h) below. (a) 0, (b) RT/2 (b) kBT/2 (c) RT (d) kBT (e) 3RT/2 (f) 3kBT/2 (g) 3RT (h) 3kBT Q4. What is expected heat capacity at a constant volume CV per one mole of CO at T = 200K (neglecting the vibrational energy)? Choose the closest. (a) 0 (b) 3RT/2 (c) 3kBT/2 (d) 3R/2 (e) 3kB/2 (f) 5RT/2 (g) 5kBT/2 (h) 5R/2 (g) 5kB/2 (h) 3RT

(i) 3kBT (j) 3R (e) 3kB Q5. A total differential of H, dH is given in the following form when P and T are independent variables:

BdTAdPdH . Which of the following statement is correct on A and B? Select ALL choices that are correct. (a) A is 0 for an ideal gas (b) B is zero for an ideal gas (c) A = (H/P)T (d) B = (H/T)P

(e) A= PT

VTV

(f) B= PT

VTV

(g) B = CV (h) B = Cp (i) Pfin

Pin

T dPTPBU ),(

3

0

2

4

6

8

10

10 20 30 40 50 60 70 80 90 100 110 120 130

V [L]

p [bar]

p1 Vi

pi V2pi Vi

0

2

4

6

8

10

10 20 30 40 50 60 70 80 90 100 110 120 130

V [L]

p [bar]

p1 Vi

pi V2pi Vi

Q6. Which are generally correct equations? Choose ALL that are correct.

(a) TT P

V

V

P

/1 (b) PHV V

H

V

P

P

H

(c) 1

VUT U

T

T

V

V

U

(d) PUV V

U

U

P

P

U

(e) HHV V

H

V

P

P

H

(f) PHV H

V

V

P

P

H

/1

Q7. Assume that the molar mass for He and Ar are 4.0 and 40, respectively. Choose the highest speed from below.

(a) <C2>1/2 for He at 300K (b) <C> for He at 300K (c)CMP for He at 300K (d) <C2>1/2 for Ar at 1500 K

(e) <C> for Ar at 1500 K (f) CMP for Ar at 1500K Q8. The graphs in the right show isotherm P-V plot for CO2 at different temperatures. Choose a all correct statements on the graph. (a) At point B, CO2 is all gas. (b) At point B, CO2 is all liquid. (c) At point D, CO2 is at a critical point. (d) At point D, CO2 is all gas. (e) At point D, CO2 is a mixture of gas and liquid. (f) At a critical point, CO2 is a mixture of gas and liquid. Q9. The P-V diagram in the right shows a closed cycle for an ideal gas. First, the ideal gas described by Ti = 300. K, Pi = 1.0 bar, and Vi = 12.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Assume that the work performed by the system is w after the entire cycle. Choose THREE correct statements below, where wc denotes the work for the total cycle. (a) wc = 0 because it forms a cycle (b) wc > 0 (c) wc < 0 (d) If the cycle is reversed, the work required is wc’ = –wC. (e) If the cycle is reversed, the work required is wc’= wC. (f) If the cycle is reversed, the work required is wc’ = 0. (g) For the work (w1) required for the initial process increasing

P = 1.0 bar to 10.0 bar, w1 > 0. (h) For the work (w1) required for the initial process increasing

P = 1.0 bar to 10.0 bar, w1 = 0. (i) For the work (w1) required for the initial process increasing P = 1.0 bar to 10.0 bar, w1 < 0.

4

0.00E+00

1.00E-03

2.00E-03

3.00E-03

0 500 1000 1500P

(C)

Speed (C) (m/s)

Distribution of Speed for O2 in Gas

Q10. Choose the most suitable sentence that represents “the first law of thermodynamics”. (a) Two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium (b) Any quantity of energy that flows across the boundary between the system and surrounding by a

force (c) Work is independent of the path between the initial and final states (d) The internal energy U of an isolated system is constant.

Q11. The diagram in the right shows distribution P of the speed (C) for O2 gas at 150 K, 300 K, 600K. What is the most probable speed for the gas at 600 K? Choose the closest. (a) 0 m/s (b) 250 m/s (c) 400 m/s

(d) 550 m/s (e) 800 m/s Q12. An ideal gas (mass m) of 1 mole is enclosed in a cubic box of length L. Assuming that each gas molecule is traveling along the x axis at a velocity of vx, and hit a wall and bounds back at a velocity of –vx, the force that the one molecule applies on one side of the wall is given by F = -p/t = 2mvx/(2L/vx) = mvx

2/L. Obtain a pressure P applied on the wall by the 1 mole of the molecule using <vx

2> is the average of vx2 and the area of one side of wall is given by L2, where M is a molar mass of

the molecule (mNA) and v2 = vx2+ vy

2 + vz2. Choose the most appropriate formula. (Hint: the question

and answer are different from the ones in Quiz 1) (a) mvx

2/L (b) m<vx2>/L (c) m<vx

2>/L3 (d) Mvx2/L (e) M<vx

2>/L (f) M<vx2>/L3

(g) m<vx2>L (h) Mvx

2L (i) mv2/3L (b) m<v2>/3L (c) m<v2>/3L3 (d) Mv2/3L

(e) M<v2>/3L3 Either one is correct.

Q13. For the following reaction, choose the most suitable formula that gives the reaction enthalpy: H2(g) + 1/2O2(g) H2O(l)

(a) ))(())(())(( 20

20

20 gHOHgOHgHH fff

(b) ))(())(()2/1())(( 20

20

20 gHOHgOHgHH fff

(c) ))(())(()2/1())(( 20

20

20 gHOHgOHgHH fff

(d) ))(())(()2/1())(( 20

20

20 gHOHgOHgHH fff

(e) ))(())(()2/1())(( 20

20

20 gHOHgOHgHH fff

5

Q14. When dz = fdx + gdy, choose all the correct conditions required for dz to be an exact differential. (Hint: there are two. Note, some equations are correct, but not related to the condition).

(a) yx

gx

fy

(b) xy y

z

x

z

(c)

xxyy y

z

yx

z

x

(d) 1

XZy z

y

y

x

x

z

(d)

xyyxx

z

yy

z

x

(e)

xy

gy

fx

(f) yx

zf

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(f) 2. Calculation/Multiple Choice Question each 4 point [ /32 points] 2.1. At a temperature 300K, the most probable speed for He was found to be 1.12 x 103 ms-1. Calculate the most probable speed for O2 at 700 K, and choose the closest value. Assume that the molar masses for He and N2 are 4.0 and 32.0. Choose the closest.

(a) 1.6 x 102 ms-1 (b) 4.2 x 102 ms-1 (c) 6.5 x 102 ms-1 (d) 8.3 x 102 ms-1 (e) 1.12 x 103 ms-1

(f) 1.90 x 103 ms-1 (g) 2.61 x 103 ms-1 (h) 9.35 x 103 ms-1

2.2-2.3 Consider a 100.0-L sample of moist air at 60°C and 1.00 atm in which the partial pressure of water vapor is 0.100 atm. Assume that dry air has the composition 78.0 mol % N2, 21.0 mol % O2, and 1.00 mol % Ar.

2.2 What are the mole percentages of H2O in the sample?

(a) 0% (b) 5% (c) 10% (d) 15% (e) 20% (f) 25% (g) 30%

2.3 The percent relative humidity is defined as %RH PH2O PH2O* where PH2O is the partial

pressure of water in the sample and PH2O

* 0.197 atm is the equilibrium vapor pressure of water at

60.°C. The gas is compressed at 60°C until the total pressure of the mixture is isothermally increased to 9.197 atm? How much fraction of H2O vapor was condensed into water after the compression?

(a) 0% (b) 20% (c) 40% (d) 60% (e) 80% (f) 100%

Vf Pf air = ViPi Vf = 100L x(0.900/9.000) =10 L nfH2O/niH2O= (PfH2OVf)/(PiH2OVi)

=0.197*10 /100*0.1 = 19.7% (remains in the air) ~80 % condenses.

2.4. The 4 lines in the right graph shows isothermal plot representing 1/V dependence of P for the same ideal gas at T = 200K, 300K, 400K, 500K. (i) Which graph of (a-d) shows that for 500 K? (ii) How much is the approximate number of moles for the molecule? (Hint use the reading of P at 1/V = 1 and R ~ 8.3 JK-1mol-1). Choose the most appropriate combination of (i, ii). (choose the closest value for ii). (a) (A, 0.1 mol) (b) (A, 0.01mol) (c) (A, 0.001 mol) (d) (D, 0.1 mol) (e) (D, 0.01mol) (f) (D, 0.001 mol)

0

20

40

60

80

0 0.5 1 1.5 2

1/V (1/m3)

P (

Pa

)

(A)

(B)

(C)

(D)

7

2.5 Obtain the most suitable expression for the total differential dV for an ideal gas when both P and T are not constant.

(a) (nR/P)dT (b) (nRT)dT (c) (nRT/P)dT

(d) (nRT/P)dP (e) (nRT)dP (f) -(nRT/P2)dP

(g) (nR/P)dT + (nRT/P2)dP (h) (nRT/P)dT - (nRT/P)dP (i) (nR/P)dT + (nRT)dP

(j) None of all

2.6 Consider an isothermal expansion of 2.0 mol of an ideal gas at 300 K from an initial volume of 3.00 L to a final volume of 6.00 L. Choose the work |w| for the process that will result in the greatest amount of work |w| being done by the system when Pext 0.0 bar. Choose the most appropriate answer.

(a) 0 (b) 2.0 mol R 300K ln(6.00L/3.00L) (c) {2.0 mol R 300K/3.00L} |6.00 L – 3.00 L|

(d) {2.0 mol R 300K/ 6.00L}|6.00 L – 3.00 L| (e)

2.7 Calculate H for the transformation of 4.00 mol of an ideal gas from 300.0 K and 1.00 atm to 600K and 17.0 atm if CV,m = 20.90 + 0.0320 (T/K) in units of JK-1mol-1. Choose the value closest to H for an ideal gas. (Hint: Cv,m is not Cp,m, & Cp,m is not Cp)

(a) 120k J (b) 53 kJ (c) 35 kJ (d) 22 kJ (e) 13.2 kJ (f) 9.4 kJ (g) 5.5 kJ (h) 0 kJ

2.8 Because (H/P)T = –CPJ–T, the change in enthalpy of a gas expanded at constant temperature can be calculated. To do so, the functional dependence of J – T on P must be known. For Ar, J – T = 3.66 x 10-6 K/Pa around 300 K. Now, 1 mol of Ar is expanded from 20 to 1.00 bar at 300. K. Then, the gas temperature was increased from 300 K to 400 K while keeping the pressure constant. Assume that µJ–T is independent of pressure and CP,m = 5/2R for Ar. Calculate H for the entire process and choose the closest value.

(a) 0 J (b) 6.5 x 102 J (c) 3.2 x 102 J (d) 2.2 x 103 J (e) 5.2 x 103 J (f) 6.5 x 104 J

J 1.44

Pa/bar x10bar2ar1 molJK2

ppμ Cdp μ C ΔH

511

ifT-Jmp,

p

p

T-Jmp,m

f

i

2

16

10

000106633185

bKPaxx ....

For the second process,

Hm = Cp,mT = 5/2R x (400 -300) = 2.08 x 103 J In total, 2.22 x 103 J.

8

Calculation Question. Choose 4 questions among the following 5 questions.

If you solve 5 without choosing, we will grade for 4 questions with the lowest scores. [ /40]

Write answers in the specified place. You should show calculations for full credit and partial credit. Please include units in calculations. 3.1 [Circle if you choose this] (Original Question P2.29) [ /10] A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless adiabatic piston. Each part contains 100.0 L of an ideal monatomic gas with CV,m = 3/2R. Initially, Ti = 400 K and Pi = 1.00 bar in each part. Heat is slowly introduced into the left part using an electrical heater until the piston has moved sufficiently to the right to result in a final pressure Pf = 8.00 bar in the right part. Consider the compression of the gas in the right part to be a reversible process.

a. Calculate the work done on the right part in this process and the final temperature in the right part.

b. Calculate the volume of the left part of the final stage

c. Calculate final temperature in the left part and the amount of heat that flowed into this part. Your Answers

(a) wR = . TR = .

(b) VL =

(c) TL= . qL = .

Show calculations for a partial credit.

The number of moles in each part is given by:

a) We first calculate the final temperature in the right side:

1

f

i

1

i

f

1

i

f

i

f

p

p

T

T

V

V

T

T

1

f

i

i

f

p

p

T

T

1

f

i

i

f

p

p

T

T

24.2

bar 50.7

bar 1.00

T

T3

53

51

i

f

fT 2.24 298 K 667 K (1 point for answer, 1 points for eq)

9

mol 02.2

K 298mol Kbar L 108.314472

L 50.0bar 1.00

T R

V p n

112-i

ii

Continued 3.1

J 109.30 K 298-K 667mol K J 8.3144722

3mol 02.2ΔT Cn ΔUw 311

V (2 point for

eq. & 1 point for answer)

b) First we calculate the volume of the right part:

-2 1 1

rfrf

rf

2.02 mol 8.314472 10 bar L K mol 667 Kn R TV 14.9 L

p 7.50 bar

Therefore L 1.85L 4.91L 1000Vlf . ( 1 point for eq. 1 point for answer)

lf lflf -2 1 1

7.50 bar 85.1 Lp VT 3800 K

n R 2.02 mol 8.314472 10 bar L K mol

(1 point for eq. 1 point for answer)

J 102.88 K 298-K 3800mol K J 8.3144722

3mol 02.2ΔT Cn ΔU 311

V

From part a) w = J 109.30 3

q = U – w = J 102.88 3 + J 109.30 3 J 105.79 3 (1 point for eq.)

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3.2 [Circle if you choose this] (Original Question P1.30) [ /10]

Using the Boltzmann distribution, the barometric pressure at the height Z above sea level in the Earth’s

atmosphere is represented as kTgZmo

iiiePP / where Pi is the partial pressure at the height z, Pi

0 is

the partial pressure of component i at sea level, g is the acceleration of gravity, k is the Boltzmann constant, T is the absolute temperature, and mi is the mass of the gas molecule (not molar mass). Consider an atmosphere that has the composition xN2 = 0.700 and xO2 = 0.300 and that T = 300. K. At the sea level, the total pressure is 1.00 bar. (a) Answer whether xN2 is increased, decreased, or constant at Z =3.0 km compared with that at the sea level. (b) Calculate PN2 and PO2 at Z = 10.0 km. (c) Calculate the mole fractions of the two components at a height of 10.0 km. Assume that the molar masses (M) of O2 and N2 are 32.00 g and 28.00 g, respectively. Use NA k= R and Mi = miNA if needed. Your Answers

(a) (b) PN2 = . PO2 = . (c) xN2= . xO2 = .

Show calculations below for a full credit. (Show equations needed for calculations.)

Use mi/kB = Mi/Ri

(a) XN2 (Z) increase as Z increases. (3 points) xN2(Z)= PN2(Z)/(PO2(Z) + PN2(Z))

= P0N2/{P0

N2 + RT

gZMM

O

NO

eP)(

022

2

} (explanation not necessary)

(b)

PaKmolJK

mmsKgmolExpPa

ePP RT

gzM

NN

N

411

32135

0

10332300318

1010819100428101

2

22

.)().(

)().().()((0.700)

PaKmolJK

mmsKgmolExpPa

ePP RT

gzM

OO

O

311

32135

0

105283003144728

1010819100032101

2

22

.)().(

)().().()((0.300)

2 points for equation including correct units. 1 point each for correct calculation (Total 4 points)

26801033210528

1052843

3

22

22 .

)..(

).(

)(

PaPa

Pa

PNPo

PoXo (1 points for eq;1 point for answer)

73201 . xx22 ON (1 point for answer)

11

3.3 [Circle if you choose this] (Original Question P3.10) [ /10]

Starting with the van der Waals equation of state for n moles of gas ( 2

2

V

an

bV

TnR

nP ), find an

expression for the total differential dP in terms of dV and dT. (a) Calculating the mixed partial derivatives (/T(P/V)T)V and (/V(P/T)V)T, Determine if P is a state function. Before solving this, clearly define the condition for dP to be an exact differential. (b) Calculate the pressure from the van der Waals equation when a =0 and b = 0.0429 L mol-1. Assume that V = 0.15 L, n = 1.00 mol, and T =300 K. Which of repulsive or attractive force is dominant in the condition for the van der Waals gas? (4 points)

(a) Condition: (/T(P/V)T)V = (/V(P/T)V)T, . (equation 2 points) Show your answers below. (Show all the equations needed for calculations.)

Van der Waals equation: V

an

bV

TnR 2

n

P

3

2

2

T

2

2

V

a 2n

bV

TnR

V

an

bV

TnR

V

nnV

P

T

bV

nR

V

an

bV

TnR

V

2

2

nnTT

P

V

2 points if both are correct;1 points if one is correct)

dT bV

nRdV

bV

TnR

V

an 2dp

23

2

nn

dTT

PdV

V

P

VT

2V

23

2

VT bV

nR

bV

TnR

V

an 2

TV

p

T nn

[1]

2

TTV bV

nR

bV

nR

VT

p

V nn

[2] (2 points if both are correct; 1 points if one is correct)

Since formula [1] = formula [2], dp is an exact differential. (1 point)

(b) When a = 0, bV

T nR

nPvdw

> V

T nR=Pideal. So a repulsive force is dominant for any conditions. (3 points)

Alternatively, you can calculate Pvdw and Pideal. Show that Pvdw > Pideal.

12

3.4 [Circle if you choose this] (P3.16) [ /10]

A mass of 36.0 g of H2O(s) at 243.0 K is dropped into 180.0 g of H2O(l) at 323.0 K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP,m(s) and CP,m(l) for H2O(s) and H2O(l) are constant throughout the temperature range of interest, respectively. Use CP,m(l) = 75.3 Jmol-1K-1 and CP,m(s) = 36.2 Jmol-1K-1. Molar fusion enthalpy (Hfusion) and molar vaporization enthalpy (Hvaporization) for H2O are 6.01 kJmol-1 and 40.65 kJmol-1, respectively. Assume that the melting point of ice is 273.0 K and that the molar mass for water is 18.0 g mol-1.

(a) Show the formula for the heat needed to increase the temperature of the ice from 243 K to 273 K.

(b) Show the equation to obtain the final temperature T (in the form of T = ..). Explain what each term represents.

(c) Calculate T.

Answer below with equations for full credit.

(a) nH2O(S) Cp,m(s) (273K - 243K) (3 points)

Or nH2O(S) Cp,m(s) (Tfin H2O(s) – Tini H2O(s))

(b)

nH2O(S) Cp,m(s) (273K - 243K) + nH2O(S)Hfusion + nH2O(S) Cp,m(l) (T - 273K)

+ nH2O(l) Cp,m(l) (T - 323K) = 0

(2 points)

T = {-nH2O(S) Cp,m(s) (273K - 243K) - nH2O(S)Hfusion + nH2O(S) Cp,m(l) (273K)

+ nH2O(l) Cp,m(l) (323K)}/{ nH2O(S) Cp,m(l) + nH2O(l) Cp,m(l)} (1 point)

The first term, the energy needed to increase the temp of ice from 243K to 273K,

The second term is the energy needed to change melt ice to liquid at 273K,

The third term is the energy needed to change melted water from 273K to T

The last term is the energy needed to change water from 323K to T.

(1 point)

(c) nH2O(s) = 36.0g/18.0 gmol-1 = 2.00 mol & nH2O(l) =180.0g/18.0 gmol-1 = 10.0 mol. (1 point)

T = {-nH2O(S) Cp,m(s) (273K - 243K) - nH2O(S)Hfusion + nH2O(S) Cp,m(l) (273K)

+ nH2O(l) Cp,m(l) (323K)}/{ nH2O(S) Cp,m(l) + nH2O(l) Cp,m(l)}

= {-2.00 mol 36.2 Jmol-1K-1 (30.0 K) – 2.00 mol x 6010 Jmol-1 + 2.00 x 75.3 Jmol-1 K-1 (273K)

+ 10.0 mol 75.3 Jmol-1 K-1 (323K)}/{ (12.0 mol)x75.3 Jmol-1K-1} =299 K (2 points)

13

3.5 [Circle if you choose this] [ /10]

(a) Calculate the molar volume and temperature of N2 at the critical point using van der Waals parameters a and b (see Table 1.1). (b) What is the ratio of the critical pressure Pc to the pressure for an ideal gas (Pideal) when T = TC and Vm = Vmc. Calculate the ratio Pc/Pideal.

a) at critical point

b3VC 27Rb

8aTC

1 1CV 3 3 0.0387 0.1161b Lmol Lmol (2 point formula; 1

point for answer & units) 2 -2

C 1 1 1

8a 8 1.370L barmolT 126.2

27Rb 27 0.0831447 0.0387K

LbarK mol Lmol

(2 point formula, 1 point for answer correct units)

b) PcVmc = (3/8)RTC (2 point)

PC = (3/8)RTC/Vmc

= (3/8)(RTC/VmC)

=(3/8) Pideal (2 points)

14

4. Derivation Questions Q1-6 2 points [ /12]

Fill a formula or equation in [Q1-Q6].

4.1 In an adiabatic expansion process, q = 0. In this case, for an isolated system, U = w. From this equation,

CVdT = -PextdV. [1]

When the process is reversible, Pext = Pint. By combining this with eq. [1], we obtain CVdT = -PintdV. For an

ideal gas, CVdT = -nRT/VdV. From this,

CV[ Q1 ]dT = -nR(1/V)dV [2]

By integrating the both sides of eq. [2], we obtain

CVln(Tfin/Tini) = [ Q2 ], [3]

where Vini and Vfin denotes the volume before and after expansion, respectively. This yields

(Tfin/Tini) = (Vfin/Vini)A,

where A = -nR/CV. This can be rewritten as the equation of P and V as

(PfinVfin/PiniVini) = (Vfin/Vini)A.

Thus, (Pfin/Pini) = (Vfin/Vini)(A – 1)

(A-1) is rewritten as [ Q3 ] using Cp/CV without the gas constant R.

[Q1: 1/T ] [Q2: -nR ln(Vfin/Vini) ] [Q3 - ]

4.2 In Joule-Thompson’s experiment (see the figure below), the left piston is under the external pressure P1, which equals the initial internal pressure. The right piston is subject to the external pressure of P2. Assuming that P1 > P2, The gas moves from the left and right while changing the volume from V1 to V2. The work applied in this process is w = wleft + wright = [ Q4 ]. Because this experiment is adiabatic, q = 0. Then, U2 – U1 = w (because U = w +q = w). This leads to H1 = H2. Here, Joule-Thompson coefficient is defined

byHH

PTJ P

T

P

T

0lim . Since 5Q

T

H

P

H

PT

= -CpT-P, this yields dH = -CpT-PdP +

CpdT. Namely, when T is constant, H = -CpT-PP assuming that Cp and T-P are constant over P. For an ideal

gas, T-P = [ Q6 ] because .0

TP

H

[Q4: -P2V2 + P1V1 ] [Q5: HP

T

]

[Q6: 0 ]

15

Calculation sheet (You can detach this sheet, if you like)

16

Equations & Constants (You can detach this sheet, if you like) Differentials dln(x)/dx = 1/x dCos(x)/dx = -Sin(x) dSin(x)/dx = Cos(x) de-x/dx = -e-x Constants

L = dm3 = 10-3 m3