Chemistry 360 Dr. Jean M. Standard

Problem Set 4 Solutions 1. Two moles of an ideal gas are compressed isothermally and reversibly at 298 K from 1 atm to 200

atm. Calculate q, w,

€

ΔU , and

€

ΔH . For an isothermal reversible expansion of an ideal gas, the work is given by

€

w = − nRT ln V2V1

#

$ %

&

' ( .

The initial and final volumes are not given. However, for an isothermal process, from the ideal gas equation we have

€

P1V1 = P2V2 , or

€

V2V1

= P1P2

.

Substituting, the work can be calculated as

€

w = − nRT ln P1P2

#

$ %

&

' (

= − 2 mol( ) 8.314 J mol−1K−1( ) 298 K( ) ln 1atm200 atm

#

$ %

&

' (

w = 26250 J .

For an ideal gas, the differential of the internal energy is given by

€

dU = CvdT and the differential of the enthalpy is given by

€

dH = CpdT . Since the process is isothermal,

€

dT = 0, and so

€

dU = 0 and

€

dH = 0. Therefore,

€

ΔU = 0 and ΔH = 0.

Finally, from the First Law of thermodynamics,

€

ΔU = q + w . Since

€

ΔU = 0 ,

€

q = − w = − 26250 J . Summary of results:

q –26250 J

w 26250 J

€

ΔU 0 J

€

ΔH 0 J

2

2. A balloon filled with 0.505 mol of gas contracts reversibly from 1.0 L to 0.10 L at a constant temperature of 5.0°C. It simultaneously loses 1270 J of heat. Calculate q, w,

€

ΔU , and

€

ΔH for the process.

For an isothermal reversible expansion of an ideal gas, the work is given by

€

w = − nRT ln V2V1

#

$ %

&

' ( .

Substituting, the work can be calculated as

€

w = − nRT ln V2V1

#

$ %

&

' (

= − 0.505mol( ) 8.314 J mol−1K−1( ) 278.15K( ) ln 0.10 L1.0 L

#

$ %

&

' (

w = 2690 J .

Even though the process described is isothermal, the overall internal energy change is not zero since the problem indicates that at the same time as the isothermal compression, heat is removed from the system. So, the First Law must be employed in order to determine

€

ΔU . The problem states that

€

q = –1270 J , so the First Law yields

€

ΔU = q+ w= −1270 J + 2690 J

ΔU = 1420 J .

Finally, to determine the enthalpy change

€

ΔH the definition of enthalpy may be used,

€

H = U + PVΔH = ΔU + Δ PV( ) .

In order to determine the change

€

Δ PV( ) = P2V2 − P1V1, the volumes are given but the pressures are not. So, the ideal gas equation may be employed to determine the initial and final pressures,

€

P1 = nRTV1

= 0.505mol( ) 0.08205L atmmol–1K–1( ) 278.15K( )

1.0 L( )P1 = 11.525atm.

and

€

P2 = nRTV2

= 0.505mol( ) 0.08205L atmmol–1K–1( ) 278.15K( )

0.1L( )P2 = 115.25atm.

3

2. Continued Now the change

€

Δ PV( ) may be determined,

€

Δ PV( ) = P2V2 − P1V1

= 115.25atm( ) 0.1L( ) − 11.525atm( ) 1.0 L( )Δ PV( ) = 0 .

It should not be a surprise that the factor

€

Δ PV( ) equals 0 since for an ideal gas another way to write this is

€

Δ PV( ) = Δ nRT( )Δ PV( ) = nR ΔT .

Since the temperature is constant,

€

ΔT = 0 and as shown above,

€

Δ PV( ) = nRΔT = 0 . The enthalpy change is therefore given by

€

ΔH = ΔU + Δ PV( )= ΔU

ΔH = 1420 J. Summary of results:

q –1270 J

w 2690 J

€

ΔU 1420 J

€

ΔH 1420 J

3. The temperature dependence of the molar constant pressure heat capacity of a real gas may be

represented by the function

€

Cp,m = α + βT + γ T 2 . where

€

α ,

€

β , and

€

γ are constants. For N2, these constants have the values

€

α = 26.984 J K−1mol−1,

€

β = 5.910×10−3 J K−2mol−1, and

€

γ = −3.377×10−7 J K−3mol−1. Determine the amount of heat required to raise the temperature of 1 mol of N2 gas from 300 to 1000 K at constant pressure. In this problem, we know that the heat at constant pressure is the enthalpy,

€

dqp = dH . The exact differential of enthalpy is

€

dH = CpdT + ∂H∂ P

#

$ %

&

' ( TdP .

4

3. Continued At constant pressure,

€

dP = 0, so the exact differential of enthalpy reduces to

€

dH = CpdT . On a molar basis, this equation becomes

€

dHm = Cp,mdT . Integrating,

€

dHmH1,m

H 2,m∫ = Cp,m dTT1

T2∫ΔHm = α + βT + γ T 2( ) dTT1

T2∫ΔHm = α T2 −T1( ) + β2 T2

2 −T12( ) + γ3 T2

3 −T13( ) .

Substituting the values of the parameters, we have

€

ΔHm = α T2 −T1( ) + β2 T22 −T1

2( ) + γ3 T23 −T1

3( )ΔHm = 26.984 J mol−1K−1( ) 1000 K − 300 K( )

+ 12 5.910×10−3 J mol−1K−2( ) 1000 K( )2 − 300 K( )2(

) * +

, -

+ 13 −3.377×10−7 J mol−1K−3( ) 1000 K( )3 − 300 K( )3(

) * +

, -

= 18889 J/mol + 2689 J/mol − 110 J/molΔHm = 21468 J/mol.

Thus, at constant pressure,

€

qp,m = ΔHm = 21468 J/mol.

4. Estimate the final temperature of one mole of gas at 200.0 atm and 19.0°C as it is forced through a porous

plug to a final pressure of 0.95 atm. The Joule-Thomson coefficient of the gas is 0.150 K/atm.

The Joule-Thomson coefficient

€

µJT is defined as

€

µJT = ∂T∂P#

$ %

&

' ( H

.

Since the problem requests an estimate of the final temperature, we can approximate the derivative in the following way,

€

µJT = ∂T∂P#

$ %

&

' ( H

≈ ΔTΔP

.

Note that in the equation above we have approximated infinitesimal changes by finite changes.

5

4. Continued We can solve the expression for the temperature difference

€

ΔT ,

€

µJT ≈ ΔTΔP

or ΔT = µJT ΔP .

Substituting the numerical values,

€

ΔT = µJT P2 − P1( )= 0.150 K/atm( ) 0.95atm − 200.0 atm( )

ΔT = − 29.86 K . The final temperature then may be determined,

€

ΔT = T2 −T1 = − 29.86 Kor T2 = T1 − 29.86K

T2 = 292.16 K − 29.86K

T2 = 262.29 K or – 10.9!C .

So for a pressure drop of nearly 200 atm, the temperature of this particular gas drops by nearly 30 degrees.

5. Determine the final temperature of 0.122 mol of an ideal monatomic gas that performs 75 J of work

adiabatically if the initial temperature is 235°C.

For an adiabatic process, q=0, so from the First Law,

€

ΔU = q + wΔU = w .

Now we are given the amount of work done and need to determine the final temperature. For an ideal gas, we can develop an expression for

€

ΔU in order to determine the temperature,

€

dU = CvdT (for an ideal gas)

dUU1

U2∫ = CvdTT1

T2∫

ΔU = Cv dTT1

T2∫ΔU = Cv ΔT .

Note that this expression requires that the constant volume heat capacity is independent of temperature in order to pull it out of the integral. So, for an adiabatic process involving an ideal gas, we have

€

ΔU = w = Cv ΔT .

6

5. Continued In order to evaluate this expression, we need the constant volume heat capacity. Any gas, monatomic, diatomic, or polyatomic, may behave ideally. Because these gases have different numbers of degrees of freedom, they also have different heat capacities. It is indicated in this problem that the ideal gas is monatomic. For such a gas, we know that the constant volume heat capacity is

€

Cv = 32 nR .

Substituting,

€

ΔU = w = Cv ΔT

= 32 nR ΔT .

We can now solve this equation for the temperature difference,

€

ΔT = 2w3nR

= 2 –75J( )

3 0.122 mol( ) 8.314 J/molK( )ΔT = − 49.29 K .

Note that the work is negative because the system performs work on the surroundings. The initial temperature is 235ºC or 508.15 K. Solving for the final temperature, we have

€

ΔT = T2 −T1 = – 49.29 Kor T2 = T1 − 49.29 K

T2 = 508.15K − 49.29 K

T2 = 458.86 K or 185.7!C .

6. A tank contains 20 L of compressed nitrogen at 10 atm and 25°C. Calculate the work when the gas is

allowed to expand reversibly and adiabatically to 1 atm pressure. Assume that the gas behaves ideally. For an adiabatic process, we know that

€

q = 0 . From the First Law,

€

ΔU = q + w . With

€

q = 0 , we have

€

w = ΔU . For an ideal gas, the differential of the internal energy is given by

€

dU = CvdT . Integrating,

€

dUU1

U2∫ = Cv dTT1

T2∫ .

Assuming that

€

Cv is independent of temperature, the integration yields

€

w = ΔU = Cv T2 −T1( ) . In order to calculate the work, the initial and final temperatures are required. However, only the initial temperature is given. The final temperature must be determined from the initial temperature, the initial volume, and the initial and final pressures in the adiabatic expansion.

7

6. Continued For a reversible adiabatic process, we derived in class that

€

Cv ln T2T1

"

# $

%

& ' = − nR ln V2

V1

"

# $

%

& ' .

Using properties of logarithms, we have

€

ln T2T1

"

# $

%

& '

Cv = ln V2

V1

"

# $

%

& '

−nR.

Taking the exponential of both sides yields

€

T2T1

"

# $

%

& '

Cv = V2

V1

"

# $

%

& '

−nR,

or

€

T2T1

"

# $

%

& '

Cv = V1

V2

"

# $

%

& '

nR.

For an ideal gas,

€

V = nRTP , so the ratio of volumes on the right hand side of the equation can be rewritten to

give

€

T2T1

"

# $

%

& '

Cv = T1P2

T2P1

"

# $

%

& '

nR.

Rearranging to get the temperatures on the left,

€

T2T1

"

# $

%

& '

Cv T2T1

"

# $

%

& '

nR = P2

P1

"

# $

%

& '

nR

T2T1

"

# $

%

& '

Cv +nR= P2

P1

"

# $

%

& '

nR

T2T1

= P2P1

"

# $

%

& '

nRCv +nR

.

Solving for the final temperature,

€

T2 = T1P2P1

"

# $

%

& '

nRCv +nR

,

T2 = T1P2P1

"

# $

%

& '

nRCp .

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6. Continued Note that nitrogen is not a monatomic ideal gas, so we need the actual heat capacity of nitrogen gas. From the CRC we have that

€

Cp,m = 29.125 J mol−1K−1. Therefore, the exponent is

€

nRC p

= nRnC p,m

= RC p,m

= 8.314 J mol−1K−1

29.125J mol−1K−1

nRC p

= 0.28546.

The final temperature is

€

T2 = T1P2P1

"

# $

%

& '

0.28546

= 298.15K( ) 1atm10 atm"

# $

%

& '

0.28546

T2 = 154.52 K .

Now, the work and internal energy change can be calculated using the relation

€

w = ΔU = Cv T2 −T1( )= nCv,m T2 −T1( ) .

Since

€

Cp = Cv + nR , in terms of molar quantities, we have

€

Cp,m = Cv,m + R . We can solve for

€

Cv,m to get

€

Cv,m = Cp,m − R

= 29.125J mol−1K−1 − 8.314 J mol−1K−1

Cv,m = 20.811J mol−1K−1.

The moles of nitrogen gas can be calculated from the initial state,

€

n = P1V1RT1

= 10 atm( ) 20 L( )

0.08205L atmmol−1K−1( ) 298.15K( )n = 8.176 mol.

Substituting,

€

w = ΔU = nCv,m T2 −T1( )

= 8.176 mol( ) 20.811J mol−1K−1( ) 154.52 K − 298.15K( )w = ΔU = − 24440 J .