Ideal Gas: PV = Nrt Or Pvm = RT

Department of Chemistry, University of Bahrain

Chemy 231

Instructor: Dr. Ali Hussain

Second Exam Time: 75 min

Name ……………………………………………… ID ………………… Sec ……

Given:

Ideal gas: PV = nRT or PVm = RT CP –CV= nR or CP,m –CV,m= R

For infinitesimal process involving only PV work:

dU = dq + dw dw = – P dV CV = (∂U/∂T)V CP = (∂H/∂T)P

dS = ( dqrev/T ) Clausius inequality: dS ≥ ( dq/T )

trsS= trsH/ Ttrs

S (Tf) = S (Ti) + ∫( Cp/T )dT

H = U + PVA = U – TSG = H – TS

wmax= A wadd,max= G

Clapeyron equation: dP/dT = ∆trsS / ∆trsV

Vm = M / ρ ( M : molar mass , ρ : density )

(∂μ/∂T)P = –Sm , (∂μ/∂P)T = Vm

I bar = 105 Pa = 0.987 atm 1 atm = 1.013105 Pa 1 atm = 760 Torr

1 m3 = 103 dm3 = 106 cm31 dm3 ≡ 1 L

1 J = 1 Pa m3 = 10–2bar dm3

R = 8.314 J K–1 mol–1

8.314 Pa m3 K–1 mol–1

0.08205 dm3 atm K–1 mol–1

0.08314 dm3 bar K–1 mol–1

62.36 dm3 Torr K–1 mol–1

Assume only PV work.

Unit of energy must be J or kJ.

Q1.(3 marks)

(i) From dH = VdP + TdS

(∂H/∂P)T = …………………………….……… + …………………………….………

= …………………………….……… + …………………………….………

[ the second expression must be in terms of directly measurable quantities ]

(ii) using the result from part (i), what is (∂H/∂P)T for an ideal gas?

show your work

Q2.(4 marks)

Consider the reaction CO (g) + CH3OH (l) → CH3COOH (l)

a) Calculate ΔrGө at 298 K from the standard enthalpies and entropies

given in the attached table (at 298 K).

b) Calculate ΔrGө at 298 K from the standard Gibbs energies of

formation given in the attached table (at 298 K).

Q3.(2 marks)

Calculate the change in Gibbs energy of 1.0 dm3 of water at 298 K when the pressure acting on it is increased from 100 kPa to 400 kPa .

Q4.(4 marks)Consider the Clapeyron equation: dP/dT = ∆trsS / ∆trsV

use T for correct statement(s) and F for false statement(s).

the Clapeyron equation can be applied to:

i)vaporization of a pure substance.[ T ]

ii)sublimation of a pure substance.[ T ]

iii)fusionof a pure substance.[ T ]

iV)the phase transition C(graphite)  C(diamond)[ T ]

Q5.(4 marks)

A pure liquid with molar mass of 46.07g/mol freezes at 269.5 K (at 1 atm) and its enthalpy of fusion is 8.68 kJ/mol.

∆(1/ρ) = 0.01899 cm3/g [ρ is the density].

What is the change in the freezing point when the pressure is increased to 987 atm?

∆T = 2.71 K

Q6.(3 marks) At 760 Torr, benzene has a normal boiling point of 80.1 oC and vapHm = 30.76 kJ/mol. If benzene is to be boiled at 35.0 oC in a vacuum distillation, to what value of P (in Torr) must the pressure be lowered? Assume vapHm is constant

over the temperature range.

Use the Clausius-Clapeyron equation:

ln(P2 /P1) = – (vapHm/ R) [(1/T2) – (1/T1)]

P = 164 Torr

Q7(3 marks)

The experimental phase diagram for carbon dioxide is shown in the graph.

a) What is the effect of increasing pressure on the melting temperature of solid

CO2?

i- melting temperature will increase

ii- melting temperature will decrease

iii- melting temperature will not change

iv- none of the above.

b) Under normal atmospheric pressure,

at what temperature can liquid

CO2 exist ?

c) What will happen to solid CO2

if it is left in the classroom now?

Q8(4 marks)

The Gibbs energy of mixing two ideal gases A and B (or two liquids A and B to form an ideal solution) at constant temperature and pressure is given by:

∆mixG = nRT(xAlnxA + xBlnxB)

n = nA+ nB ; xA : mole fraction of A ; xB : mole fraction of B

(a) what is the driving force of mixing perfect gases?

(b) what is the driving force of mixing two liquids to form ideal solution?

And correct the false statement(s)

-in both perfect gases and ideal solutions, there are no interaction between the molecules. [ F ]

-in real solutions, both ∆mixHand ∆mixS can be positive or negative. [ T ]