Tutorial 2 Memo A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate PDF

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CMY282 – PHYSICAL CHEMISTRYTutorial 2: The First Law of ThermodynamicsQUESTION 1At 298 K the C V ,mvalue of SO 2is greater than that of CO 2. At very high temperatures (> 1000 K) the C V ,mof CO 2is greaterthan that of SO 2. Explain why these differences are observed.Solution. At high temperature...

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CMY282 – PHYSICAL CHEMISTRY Tutorial 2: The First Law of Thermodynamics QUESTION 1 At 298 K the CV,m value of SO2 is greater than that of CO2. At very high temperatures (> 1000 K) the CV,m of CO2 is greater than that of SO2. Explain why these differences are observed. Solution. At high temperatures, 𝑘𝐵 𝑇 becomes large enough that the vibrational degrees of freedom of a molecule can fully contribute to the heat capacity. CO2 is a linear molecule with three (3) translational, two (2) rotational and four (3×3 − 5 = 4) vibrational degrees of freedom. Remember that we have 3𝑁 − 5 modes of vibration for linear molecules. Recalling that translational and rotational degrees of freedom each contributes

degree of freedom contributes 𝑅, we have that

𝑅 𝑅 𝐶𝑉,mCO2 = 3 ( ) +2 ( ⏟ ) + 4𝑅 ⏟2 ⏟2 Translation

Rotation

Vibration

=

𝑅

2

to 𝐶𝑉,m , while each vibrational

13 𝑅 2

SO2, on the other hand, is not linear. It has three (3) translational and three (3) rotational degrees of freedom and three (3×3 – 6=3) vibrational degrees of freedom. Remember that we have 3𝑁 − 6 modes of vibration for linear molecules. Therefore, 𝑅 𝑅 ⏟ ) + 3𝑅 ) +3 ( = 6𝑅 𝐶𝑉,mSO2 = 3 ( ⏟2 ⏟2 Vibration Translation

Rotation

We see that in the linear molecule the substitution of an extra vibrational degree of freedom for the absent rotational degree of freedom leads to a larger 𝐶𝑉,m , since vibrational degrees of freedom contribute twice as much to the heat capacity. At room temperature, however, the vibrational degrees of freedom contribute only slightly and the extra rotational degree of freedom for the non-linear molecule gives it the larger 𝐶𝑉,m . We can summarise the findings as follows 𝑪𝑽,𝐦𝐂𝐎

𝑪𝑽,𝐦𝐒𝐎

Comparison of 𝑪𝑽,𝐦 Values

Low T (Translational + Rotational)

5 𝑅 2

3𝑅

SO2 > CO2

High T (Translational + Rotational + Vibrational)

13 𝑅 2

6𝑅

CO2 > SO2

𝟐

CMY 282 – PHYSICAL CHEMSITRY

𝟐

Tutorial 2: The First Law of Thermodynamics

Page 1 of 11

QUESTION 2 A sample consisting of 2.00 mol He is expanded isothermally at 22°C from 22.8 dm3 to 31.7 dm3. (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure). For the three processes calculate q, w, ΔU, and ΔH. Solution. For a perfect gas under isothermal conditions, we know that Δ𝑈 = 0, since the change in internal energy depends solely on temperature, when the temperature is constant this quantity must also remain constant. In addition, we know that Δ𝐻 = Δ𝑈 + Δ(𝑝𝑉) = Δ𝑈 + Δ(𝑛𝑅𝑇), therefore Δ𝐻 = 0. Hence, for all three cases (a), (b) and (c) this will be the case. (a) Δ𝑈 = Δ𝐻 = 0. For reversible, isothermal expansions, the work is given by 𝑤 = −𝑛𝑅𝑇 ln (

𝑉𝑓 31.7 dm3 ) = −(2.00 mol)(8.314 J ⋅ K −1⋅ mol−1) ln ( ) = −1.62 × 103 J = −1.62 kJ 𝑉𝑖 22.8 dm3

Then from the First Law of Thermodynamics Δ𝑈 = 𝑞 + 𝑤 = 0, and so 𝑞 = −𝑤 = + 1.62 kJ.

(b) Δ𝑈 = Δ𝐻 = 0. When expansion is done against a constant external pressure, the work done by the gas is 𝑤 = −𝑝ex Δ𝑉

However, we first need the final pressure of the gas. This can be calculated using the perfect gas law 𝑝𝑓 𝑉𝑓 = 𝑛𝑅𝑇

𝑝ex= 𝑝𝑓 =

(2.00 mol)(0.08206 L ⋅ atm ⋅ K −1⋅ mol−1)(22 + 273 K) = 1.53 atm 31.7 L

Then we can calculate the work done by the gas 𝑤 = −𝑝ex Δ𝑉 = −(1.53 atm)(31.7 L − 22.8 L) |101.3

J | = −1.38 × 103 J = −1.38 kJ L ⋅ atm

Then, again using the First Law of Thermodynamics we can calculate the heat absorbed 𝑞 = −𝑤 = +1.38 kJ (c)

Δ𝑈 = Δ𝐻 = 0. For a free expansion, 𝑤 = 0, and combining this fact with the First Law of Thermodynamics we obtain that 𝑞 = −𝑤 = 0.

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 2 of 11

QUESTION 3 The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression 𝐶𝑝 /(J K–1 ) = 20.17 + 0.4001(𝑇/K). Calculate q, w, ΔU, and ΔH when the temperature is raised from 0°C to 100°C (a) at constant pressure, (b) at constant volume.

Solution. Assume we have 1.00 mol of gas. (a) At constant pressure we know that 𝑞𝑝 = Δ𝐻. So, in order to calculate Δ𝐻, we must recall the definition of 𝐶𝑝 as follows 𝐶𝑝 = (

𝜕𝐻

𝜕𝑇

)

𝑝

Then, using the information provided, we can calculate the change in enthalpy by integration of 𝐶𝑝 as follows 𝑞𝑝 = Δ𝐻

𝑇2

= ∫ 𝐶𝑝 𝑑𝑇 𝑇1

373 K

= ∫

273 K

(20.17 + 0.4001𝑇)𝑑𝑇

373 K 1 = [20.17𝑇 + (0.4001)𝑇 2 ] 2 273 K

1 = 20.17(373 − 273) + (0.4001)(3732 − 2732 ) 2 = 14.9 × 103 J = 14.9 kJ . Next, we can calculate the work done by the gas as follows 𝑤 = −𝑝ex Δ𝑉 = −𝑛𝑅Δ𝑇 = −(1.00 mol)(8.314 J ⋅ K −1⋅ mol−1)(100 K) = −831 J So, finally the internal energy can be calculated by the First Law of Thermodynamics Δ𝑈 = 𝑞 + 𝑤 = (14.9 × 103 J) + (−831 J) = 14.1 kJ (b) The internal energy and enthalpy of a perfect gas depend only on the temperature, therefore at constant volume we recall our values from part (a) as follows Δ𝐻 = 14.9 kJ

and

Δ𝑈 = 14.1 kJ

However, at constant volume we also know that w=0 since 𝑤 = −𝑝ex 𝛥𝑉 and 𝛥𝑉 = 0. Which is always the case at constant volume. Then from the First Law of Thermodynamics 𝑞𝑉 = Δ𝑈 = 14.1 kJ

Note that at constant volume we will always have that 𝑞𝑉 = Δ𝑈.

This example is irreversible (there is non-expansion work present, 𝑉Δ𝑝), if it were reversible, then Δ𝐻 = Δ𝑈 = 𝑞𝑉 !

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 3 of 11

QUESTION 4 When 2.0 mol CO2 is heated at a constant pressure of 1.25 atm, its temperature increases from 250 K to 277 K. Given that the molar heat capacity of CO2 at constant pressure is 37.11 J⋅K–1⋅mol–1, calculate q, ΔH, and ΔU. Solution. At constant pressure we know that the heat and the enthalpy change will be equal, and then assuming that 𝐶𝑝,m is constant over the given temperature range (i.e. we need not integrate) we can use the following expression Δ𝐻 = 𝑞𝑝

= 𝑛𝐶𝑝,m Δ𝑇

= (2.0 mol)(37.11 J ⋅ K−1⋅ mol−1 )(277 K − 250 K) = 2.00 × 103 J = 2.00 kJ

Next, from the definition of enthalpy we can obtain the internal energy. Since Δ𝐻 = Δ𝑈 + 𝑛𝑅Δ𝑇, we can rewrite the equation with Δ𝑈 as the subject, as follows Δ𝑈 = Δ𝐻 − 𝑛𝑅Δ𝑇

= 2.00 × 103 J − (2.0 mol)(8.314 J ⋅ K −1⋅ mol−1)(277 K − 250 K) = 1.55 × 103 J = 1.55 kJ

Alternatively, we could calculate the internal energy using the expression Δ𝑈 = 𝑛𝐶𝑉,m Δ𝑇, where

𝐶𝑉,m= 𝐶𝑝,m− 𝑅 = 37.11 J ⋅ K −1⋅ mol−1− 8.314 J ⋅ K −1⋅ mol−1= 28.796 J ⋅ K −1⋅ mol−1

Then Δ𝑈 = 𝑛𝐶𝑉,m Δ𝑇 = (2.0 mol)(28.796 J ⋅ K −1⋅ mol−1)(277 K − 250 K) = 1.55 kJ which is the same answer as previously obtained.

How would we have calculated the work? (This was not part of the question) Well, since it is a constant pressure process, we would use 𝑤 = −𝑝𝑒𝑥 𝛥𝑉 = −𝑛𝑅𝛥𝑇 = −449 J or we could have used the First Law of Thermodynamics 𝑤 = 𝛥𝑈 − 𝑞𝑝 = −450 J

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 4 of 11

QUESTION 5 A sample of 5.0 mol CO2 is originally confined in 15 dm3 at 280 K and then undergoes adiabatic expansion against a constant pressure of 78.5 kPa until the volume has increased by a factor of 4.0. Calculate q, w, ΔT, ΔU, and ΔH. (The final pressure of the gas is not necessarily 78.5 kPa.)

Solution. First, we can calculate the final and the change in volume by the given relation 𝑉𝑖 = 15 dm3 ∴ 𝑉𝑓 = 4 × 15 = 60 dm3

⇒

Δ𝑉 = 45 dm3

We know that for an adiabatic process, the heat is zero per definition 𝑞 = 0. Next, since this is an expansion process against a constant external pressure, the work can be calculated as follows (we have two methods, each using a different set of units) 𝑤

= −𝑝ex Δ𝑉 = − (78.5 × 103 Pa) (45 dm 3 | = −3.5 × 103 J

1 m3 |) (10 dm)3

𝑤

= −𝑝ex Δ𝑉 = − (78.5 kPa | = −3.5 × 103 J

= −3.5 kJ

101.3J 1atm |) (45L | |) 101.325 kPa L ⋅ atm

= −3.5 kJ

Then from the First Law of Thermodynamics we can calculate the change in internal energy as follows (in Atkins’ there is an expression 𝑤ad= Δ𝑈, which just means that adiabatic work equals the change in internal energy as we do here) Δ𝑈 = 𝑞 + 𝑤 = 𝑤 = −3.5 kJ

We can also relate the adiabatic work and then in essence the internal energy to Δ𝑇 as follows 𝑤ad= Δ𝑈 = 𝐶𝑉 Δ𝑇 = 𝑛(𝐶𝑝,m− 𝑅)Δ𝑇

⇒

Δ𝑇 =

𝑤

𝑛(𝐶𝑝,m− 𝑅)

Where 𝐶𝑝,m is obtained from the Resource section of your Textbook. Therefore, the change in temperature is Δ𝑇 =

−3.5 × 103 J = −24 K (5.0 mol)(37.11 − 8.3145) J ⋅ K −1⋅ mol−1

Next, assuming we have a perfect gas (i.e. a gas at low pressure), we can calculate the change in enthalpy as follows Δ𝐻 = Δ𝑈 + Δ(𝑝𝑉) = Δ𝑈 + 𝑛𝑅Δ𝑇 = −3.5 × 103 J + (5.0 mol)(8.3145 J ⋅ K −1⋅ mol−1)(−24 K) = −4.5 × 103 J Alternatively, we could have calculated the enthalpy using 𝛥𝐻 = 𝑛𝐶𝑝,𝑚 𝛥𝑇, since 𝛥𝑈 and 𝛥𝐻 are state functions. WHY is it mentioned that the final pressure of the gas is not necessarily 78.5 kPa? The reason is that the pressure changes which complicates matters (𝑉𝛥𝑝 can be viewed as non-expansion work), so since 𝛥𝐻 = 𝛥𝑈 + 𝛥(𝑝𝑉) = 𝛥𝑈 + 𝑝𝛥𝑉 + 𝑉𝛥𝑝, we can “bypass” this problem by using 𝛥𝐻 = 𝛥𝑈 + 𝑛𝑅𝛥𝑇 as we did.

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 5 of 11

QUESTION 6 3

A sample consisting of 1 mol of perfect gas atoms (for which CV,m = 2R) is taken through the cycle shown in Fig. 2.34.

(a) Determine the temperature at the points 1, 2, and 3. (b) Calculate q, w, ΔU, and ΔH for each step and for the overall cycle.

Solution. In all steps in the problem we assume them to be reversible. We first analyze the figure as follows Constant Pressure Constant Volume (a) The temperatures are readily obtained from the perfect gas equation, 𝑇 = ends of an isotherm, they have the same value as follows 𝑇1 = 𝑇3 =

𝑝𝑉

, so since 𝑇1 and 𝑇3 are points on the

𝑛𝑅

(1.00 atm)(22.4 dm3 ) = 273 K (1.00 mol)(0.0821 0.08206 dm3 ⋅ atm ⋅ K−1⋅ mol−1 )

Similarly, 𝑇2 =

(1.00 atm)(44.88 dm3 ) = 546 K (1.00 mol)(0.0821 0.08206 dm3 ⋅ atm ⋅ K−1⋅ mol−1 )

In the following Table we summarize the total cycle of easier use. Total Cycle

CMY 282 – PHYSICAL CHEMSITRY

State

𝑝/atm

𝑉/dm3

𝑇/K

1

1.00

22.44

273

2

1.00

44.88

546

3

0.50

44.88

273

Tutorial 2: The First Law of Thermodynamics

Page 6 of 11

(b) We will break this part of the question up into three different sections, as follows Step 1→2 This step takes place at constant pressure, with a volume and temperature change. Hence, we can calculate the work as follows 𝑤 = −𝑝ex Δ𝑉 = −𝑝Δ𝑉 = −𝑛𝑅Δ𝑇 = −(1.00 mol)(8.3145 J ⋅ K−1⋅ mol−1)(546 − 273)K = −2.27 × 103 J Next, we can calculate the change in internal energy as follows 3 Δ𝑈 = 𝑛𝐶𝑉,m Δ𝑇 = (1.00 mol) ( × 8.3145 J ⋅ K−1⋅ mol−1 ) (273 K) = +3.40 × 103 J 2 The heat can be calculated from the First Law of Thermodynamics 𝑞 = Δ𝑈 − 𝑤 = 3.40 × 103 J − (−2.27 × 103 J) = +5.67 × 103 J We also know that at a constant pressure that Δ𝐻 = 𝑞𝑝 = +5.67 × 103 J Alternatively, we could calculate the change in enthalpy using Δ𝐻 = 𝑛𝐶𝑝,m Δ𝑇.

If this step were irreversible, then 𝑤, 𝑞 and Δ𝐻 would be indeterminate.

Step 2→3 This step is a constant volume process, hence 𝑤 = 0. Then, we know that at constant volume 𝑞𝑉 = Δ𝑈 3 𝑞𝑉 = Δ𝑈 = 𝑛𝐶𝑉,m Δ𝑇 = (1.00 mol) ( × 8.3145 J ⋅ K−1⋅ mol−1 ) (−273 K) = −3.40 × 103 J 2

For the change in enthalpy we consider Δ𝐻 = Δ𝑈 + Δ(𝑝𝑉) = Δ𝑈 + 𝑛𝑅Δ𝑇 for a perfect gas. So,

Δ𝐻 = Δ𝑈 + 𝑛𝑅Δ𝑇 = −3.40 × 103 J + (1.00 mol)(8.3145 J ⋅ K −1⋅ mol−1)(−273 K) = −5.67 × 103 J

Alternatively, we could have argued that Δ𝑇1→2= −Δ𝑇2→3 and so it follows that Δ𝑈1→2= −Δ𝑈2→3 𝑎𝑛𝑑 Δ𝐻1→2= −Δ𝐻2→3

Step 3→1 For any isothermal process Δ𝑈 = Δ𝐻 = 0. Since they are state functions and we have an ideal gas.

Then, from the First Law of Thermodynamics, we obtain the expression 𝑞 = −𝑤 . Also, for an ideal gas under reversible conditions, the work done by the gas can be expressed as 𝑤 = −𝑛𝑅𝑇 ln ( have that −𝑞 = 𝑤 = −𝑛𝑅𝑇 ln (

𝑉𝑓

𝑉𝑖

), so for our step 3→1, we

𝑉1 22.4 dm3 ) = +1.57 × 103 J ) = −(1.00 mol)(8.3145 J ⋅ K −1⋅ mol−1)(273 K) ln ( 44.8 dm3 𝑉3

and so 𝑞 = −1.57 × 103 J.

If this step were not reversible, then 𝑞 and 𝑤 would have different values, which would be determined by the details of the process.

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 7 of 11

We summarize the findings of Q6 (b) in the following table. Thermodynamic quantities calculated for reversible steps 𝒒/kJ

𝒘/kJ

𝚫𝑼/kJ

𝚫𝑯/kJ

+5.67

–2.27

+3.40

+5.67

𝑉 constant

–3.40

0

–3.40

–5.67

Isothermal, reversible

–1.57

+1.57

0

0

+0.70

–0.70

0

0

Step

Process

1→ 2

𝑝 constant = 𝑝ex

2→3 3→1 Cycle

Comment. All values can be determined unambiguously for the reversible cycle. The net result of the cycle process is that 700 J of heat has been converted to work!

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 8 of 11

QUESTION 7 A sample consisting of 2.0 mol CO2 occupies a fixed volume of 15.0 dm3 at 300 K. When it is supplied with 2.35 kJ of energy as heat its temperature increases to 341 K. Assume that CO2 is described by the van der Waals equation of state, and calculate w, ΔU, and ΔH. Solution. Since the volume is fixed the work done by the gas during the process is zero, 𝑤 = 0. Moreover, since we have a constant volume process, we know that Δ𝑈 = 𝑞𝑉 , and 𝑞𝑉 is given as 2.35 kJ, so Δ𝑈 = 𝑞𝑉 = 2.35 kJ. For the change in enthalpy we must work a bit harder. First, we consider the expression for change in enthalpy Δ𝐻 = Δ𝑈 + Δ(𝑝𝑉) = Δ𝑈 + 𝑉Δ𝑝 + 𝑝Δ𝑉 = Δ𝑈 + 𝑉Δ𝑝

[Δ𝑉 = 0, so pΔV = 0]

Previously we would just have said but Δ(𝑝𝑉) = 𝑛𝑅Δ𝑇, but this is only valid for an ideal gas! So, we need to calculate the change in pressure, and the gas in question is a van der Waals gas, so the pressure of the gas and the change is 𝑝=

𝑎 𝑅𝑇 − 2 𝑉m − 𝑏 𝑉m

⇒

Δ𝑝 =

𝑅Δ𝑇 𝑉m − 𝑏

The only parameter that has a non-zero change is the temperature, since at constant volume Δ𝑉𝑚 = 0 and the other components are all constants hence their changes are also zero. Therefore, Δ𝐻 = Δ𝑈 + 𝑉Δ𝑝 = Δ𝑈 + 𝑉 (

𝑅Δ𝑇 ) 𝑉m − 𝑏

We have Δ𝑈, so we just need 𝑉m ,𝑅,Δ𝑇 and 𝑏. In the Resource section of Atkins’ 𝑏 = 4.3 × 10−2 dm3 ⋅ mol−1, and we know that 𝑅 = 8.3145 J ⋅ K−1⋅ mol−1. The change in temperature is just Δ𝑇 = 341 − 300 = 41 K. To calculate the molar volume of the gas we use the information as follows. If 2.0 mol gas occupies 15.0 dm3, then the molar volume is simply 𝑉m =

15.0 dm3 = 7.5 dm3 ⋅ mol−1 2.0 mol

Now that we have all the parameters required, we can calculate the 𝑉Δ𝑝 term as follows

(8.3145 J ⋅ K−1⋅ mol−1)(41 𝐾) 𝑅Δ𝑇 ) = 0.68 kJ ) = (15.0 𝑑𝑚3 ) ( 𝑉Δ𝑝 = 𝑉 ( (7.5 − 4.3 × 10−2) dm3 ⋅ mol−1 𝑉m − 𝑏

Therefore, the change in enthalpy is Δ𝐻 = Δ𝑈 + 𝑉 (

𝑅Δ𝑇

𝑉m − 𝑏

) = 2.35 kJ + 0.68 kJ = +3.03 kJ

Alternatively, if this involved too much mathematics for you, you could calculate 𝑝𝑖 and 𝑝𝑓 from the van der Waals equation and substitute it in 𝑉𝛥𝑝 = 𝑉 (𝑝𝑓 − 𝑝𝑖 ) and calculate the change in enthalpy from that.

What if we assumed ideal gas behavior? Well, then we would have simply said 𝛥𝐻 = 𝛥𝑈 + 𝛥(𝑝𝑉) = 𝛥𝑈 + 𝑛𝑅𝛥𝑇, and we would get the answer 𝛥𝐻ideal= 3.03 kJ. Which is interesting right? We found that 𝛥𝐻ideal= 𝛥𝐻vdW and so we can conclude that CO2 behaves ideally in this temperature range, with the given molar volume.

CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

Page 9 of 11

QUESTION 8 7

A sample of 1.00 mol perfect gas molecules with Cp,m = 2R is put through the following cycle: (a) Constant volume heating to twice its initial pressure, (b) Reversible, adiabatic expansion back to its initial temperature, (c) reversible isothermal compression back to 1.00 atm. Calculate q, w, ΔU, and ΔH for each step and overall. Assume the initial temperature to be 298 K. Solution. Since we have three processes, subscripts 1, 2 and 3 refer to (a), (b) and (c) respectively. (a) At constant volume we know that 𝑤 = −𝑝ex Δ𝑉 = 0. Next, we assume that 𝑇1 = 298 K, and since we have a perfect gas, we know that 𝑝1 𝑇1 = 𝑝2 𝑇2

Next, remember that 𝐶𝑉,m= 𝐶𝑝,m− 𝑅 =

5

2

∴ 𝑇𝑓 = 2 × 298 K

𝑅. Then, the change in internal energy can be calculated using

5 Δ𝑈 = 𝑛𝐶𝑉,m Δ𝑇 = (1.00 𝑚𝑜𝑙) ( × 8.3145 J ⋅ K−1⋅ mol−1 ) (298 K) = 6.19 × 103 J = 6.19 kJ 2

Then from the First Law of Thermodynamics we have that Δ𝑈 = 𝑞𝑉 = 6.19 kJ Finally, the change in enthalpy can be calculated using the relation (for a perfect gas we may do the following) Δ𝐻 = Δ𝑈 + Δ(𝑝𝑉) = Δ𝑈 + 𝑛𝑅Δ𝑇 = 6.19 kJ + (1.00 𝑚𝑜𝑙)(8.3145 J ⋅ K−1⋅ mol−1)(298 K) = 8.67 kJ

Alternatively, we could have used 𝛥𝐻 = 𝑛𝐶𝑝,𝑚 𝛥𝑇.

(b) Per definition for an adiabatic process, 𝑞 = 0. Because the change in internal energy and the change in enthalpy are only dependent on the temperature, and here we have that Δ𝑇(𝐚)= −Δ𝑇(𝐛), (the subscripts refer to the processes (a) and (b)) we can immediately say that Δ𝑈(𝐛)= −Δ𝑈(𝐚)= −6.19 kJ

and

ΔH(𝐛)= −Δ𝐻(𝐚)= −8.67 kJ

Then, the work for an adiabatic process, by the First Law of Thermodynamics states that 𝑤ad= Δ𝑈 = −6.19 𝑘𝐽. (c) For an isothermal compression, Δ𝑇 = 0. Then since both the change in enthalpy and change in internal energy only always depend on the temperature, they are both zero Δ𝑈 = Δ𝐻 = 0. Then by the First Law of Thermodynamics 𝑞 = −𝑤. We can calculate the work for a reversible process as follows 𝑤 = −𝑛𝑅𝑇1 ln (

𝑉1

𝑉3

)

Where 𝑇1 = 298 K. However, we need to calculate 𝑉1 and 𝑉3 , we can do this by using the ideal gas law for 𝑉2 𝑉1 = 𝑉2 =

𝑛𝑅𝑇1 (1.00 mol)(0.08206 L ⋅ atm ⋅ K−1⋅ mol−1)(298 K) = 24.45 L = 1.00 atm 𝑝1

and for 𝑉3 we use the adiabat equation as follows (remember 𝑐 =

𝐶𝑉,m 𝑅

5

5

= 2)

2 × 298 K 2 𝑇2 𝑐 ) = 138.3 L 𝑉3 = 𝑉2 ( ) = 24.45 L ( 𝑇3 298 K CMY 282 – PHYSICAL CHEMSITRY

Tutorial 2: The First Law of Thermodynamics

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Finally, then the work is 𝑤 = −𝑛𝑅𝑇1 ln (

𝑉1

24.45 ) = + 4.29 × 103 J ) = −(1.00 𝑚𝑜𝑙)(8.314 J ⋅ K−1⋅ mol−1)(298 K) ln ( 138.3

𝑉3 So, 𝑤 = + 4.29 kJ, and then 𝑞 = ...