Exam 1 Solutions 100 points

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1 Chemistry 360 Fall 018 Dr. Jean M. Standard September 19, 018 Name KEY Exam 1 Solutions 100 points 1.) (14 points) A chunk of gold metal weighing g at 800 K is dropped into g of liquid water at 300 K in an insulated container at 1 bar pressure. Assume that the molar constant pressure heat capacities of Au and H O are independent of temperature. [Au g/mol; H O 18.0 g/mol; C p,m ( Au) 5.4 J mol 1 K 1 ; C p,m ( H O) 75.9 J mol 1 K 1 ] a.) Calculate the temperature of the system once thermal equilibrium has been reached. Since the process occurs in an insulated container, there is no heat lost or gained in the process, q 0. This can be broken up into the heat given off by the gold and the heat absorbed by the water, q Au + q H O 0. Now, since the system is at constant pressure, the heat evolved or absorbed is equal to the enthalpy change, q p ΔH C p ΔT. Note that this equation is valid for constant heat capacities as was specified above. Substituting, q Au + q H O 0 C p,au ( T f T i,au ) + C p,h O ( T f T i,h O ) 0. Writing this expression in terms of molar heat capacities gives n Au C p,m,au ( T f T i,au ) + n H OC p,m,h O ( T f T i,h O ) 0. All the quantities in this expression are known except the final temperature. Solving for the final temperature gives T f n Au C p,m,au T i,au + n H O C p,m,h O T i,h O n Au C p,m,au + n H O C p,m,h O T f 315 K. 800 K ( mol) 5.4 Jmol 1 K 1 ( mol) 5.4 Jmol 1 K 1 + ( mol ) 75.9 Jmol 1 K mol 300 K 75.9 Jmol 1 K 1 b.) What is the expected enthalpy change, ΔH, for the process? Explain without performing any calculations. Since the overall q for the process equals 0 because the process takes place in an insulated container, and the process also occurs at constant pressure, we have ΔH q p 0.

2 .) (14 points) Sketch a graph illustrating the typical behavior of the compression factor Z as a function of pressure for a real gas. You should show two curves on your graph: one for low temperature (something like 00 K) and one for high temperature (something like 1000 K) for a real gas such as diatomic nitrogen. Make sure that you label which is which and also indicate the ideal gas value of the compression factor on the graph. Then, briefly explain: (a) the physical origin of negative deviations from ideality; (b) the physical origin of positive deviations from ideality; and (c) whether a typical real gas is expected to behave ideally over a wider range of pressures at low or high temperature. A typical plot of the compression factor Z vs. the pressure P for a real gas such as N at various temperatures is shown below. Note that only two temperatures have to be shown (low and high). The ideal gas result, Z1, is indicated by the dashed line on the graph. Note that the scale of P on the x-axis usually ranges from P0 to P1000 atm or so. (a) The region of negative deviation for a real gas typically occurs at low to moderate pressures. The negative deviations (i.e., Z values < 1) are a result of intermolecular interactions. At low pressure, the atoms or molecules of a real gas interact with one another and "stick together". This results in a pressure that is lower than expected, and therefore a compressibility factor Z that is less than 1. Note however, that in the limit of very low pressures (as P 0 ), Z values for a real gas approach 1 as the gas molecules become too sparse for intermolecular interactions to significantly impact the pressure. (b) The region of positive deviation for a real gas occurs at high pressure. The positive deviations (i.e., Z values <>1) are a result of the finite size of real gases. An ideal gas takes up no space, but for a real gas, as the pressure increases, the atoms or molecules get crushed together and cannot get any closer. This leads to a volume that is larger than that for the ideal gas case, and therefore a compressibility factor Z that is greater than 1. (c) As the temperature increases, the strength of the intermolecular interactions becomes much smaller than the thermal energy the molecules possess, so the intermolecular interactions play a much smaller role in determining the pressure of the gas. Thus, at higher temperatures (as illustrated above), the negative deviations decrease and the gas behaves ideally over a larger range of pressure. [We also observed this when studying the isotherms of a real gas, which were much more like ideal gas isotherms at high temperatures.]

3 3 3.) (14 points) The surface tension of a liquid is defined as the propensity of the liquid to minimize its surface area due to the intermolecular interactions between molecules at the surface. An equation of state which describes the surface tension τ of a liquid, measured in units of J/m or N/m, is given by τ + b A A c. Here, τ is the surface tension, A is the area of the surface, T is temperature, R is the gas constant, and b and c are constants for the specific liquid. The infinitesimal work done in changing the area of a liquid surface is defined to be dw τ da. Using the equation of state, derive an expression for the work done in a reversible isothermal expansion that increases the surface of the liquid from area A 1 to area A at temperature T. We can obtain an expression for the surface tension τ from the equation of state given above. Solving the equation for τ yields τ A c b A. Substituting this relation into the equation for work yields w A A c b da, A 1 A or w A A c da A 1 A b A da. A 1 Since the process is isothermal, T can be pulled out of the integrals along with the other constants to give A 1 w A c da b A 1 A 1 A da. A 1 Integrating leads to the result w ln A c A 1 c 1 + b 1. A A 1

4 4 4.) (15 points) True/false, short answer, multiple choice. a.) True or False : The Zeroth Law of thermodynamics provides a statement about chemical equilibrium in molecular systems. b.) True or False: A typical rubber band contracts when heated under constant tension conditions. c.) Short answer An intensive variable is independent of the size of the system. d.) Short answer The critical point is the location on a pressure-volume diagram where condensation first occurs as the system is cooled from a gaseous state at high temperature. e.) Multiple Choice. The Equipartition Theorem predicts that the molar internal energy of an ideal monatomic gas has the following form: 1) U m 5. ) U m 3. 3) U m 5 R. 4) U m 3 R.

5 5.) (14 points) A physical chemist has obtained a sample of ethane gas (C 3 H 6 ) at a temperature of 300 K and a pressure of 1.0 bar. a.) Assuming that ethane behaves as an ideal gas under the conditions given above, calculate the expected molar volume of the sample in L/mol. Assuming that the gas obeys the ideal gas equation, we have for the molar volume V m P ( L bar mol -1 K -1 ) 300 K 1.0 bar V m 4.9 L/mol. 5 b.) The physical chemist measured experimentally the molar volume of the sample of ethane under the conditions given above and found it to be.0 L/mol. Assuming that the behavior of ethane may be described by the Virial Equation, determine the value of the Second Virial Coefficient B for ethane. You may set the Third Virial Coefficient, C, equal to zero. Treating the gas as obeying the Virial Equation, we have Z PV m 1 + B V m + C V m. where B and C are the Second and Third Virial Coefficients, respectively. Setting C0, we have Z PV m 1 + B V m. The expression can be solved for B (by multiplying both sides by V m and rearranging), Substituting, B PV m V. m B PV m V m ( 1.0 bar).0 L/mol L bar mol -1 K K B.6 L/mol. (.0 L/mol)

6 6 5. Continued c.) What is the primary physical reason for the difference in molar volume observed for the ideal gas and the real gas in parts (a) and (b)? For an ideal gas, we know that the compression factor equals 1, or Z1. For the real gas in part (b), the compression factor is Z PV m ( 1.0 bar).0 L/mol L bar mol -1 K -1 Z K We see that the compression factor Z calculated for the real gas is in this case lower than the ideal gas value. This corresponds to the region where attractive intermolecular interactions lead to deviations from ideality.

7 6.) (14 points) Determine the quantities q, w, ΔU, and ΔH for the following processes involving two moles of an ideal monatomic gas. Express your results in Joules. a.) Two moles of neon gas initially at 400ºC and.0 bar are expanded reversibly and isothermally to a final pressure of 0.5 bar. Starting from the basic definition of work, we have 7 w P ext dv This process is reversible. Therefore, P ext P. This yields Using the ideal gas equation, we have P n V V w P dv.. Substituting, w V n V dv. Since the process is isothermal, we can pull out T along with the other constants and integrate the remaining expression. w n V 1 V dv $ w n ln V ' & ). % ( We need to determine the initial and final volumes. The ideal gas equation can be utilized for this purpose. The initial volume is n 1 P K ( mol) L bar mol 1 K 1 (.0 bar) 56.0L. Noting that the initial and final temperatures are the same, the final volume is V n P K ( mol) L bar mol 1 K 1 ( 0.5bar) V 447.7L.

8 8 6 a.) Continued The work in Joules can then be determined by substitution, w n ln V ( mol) ( J mol 1 K 1 ) K w 370 J. ln 447.7L 56.0 L To determine the change in internal energy, we know that for an ideal gas, du C v dt, or in integrated form (given that C v is independent of temperature), ΔU C v ΔT. Since the process is isothermal, ΔT 0, and therefore we know that ΔU 0. To determine the change in enthalpy, we know that for an ideal gas, dh C p dt, or in integrated form (given that C p is independent of temperature), ΔH C p ΔT. Since the process is isothermal, ΔT 0, and therefore we know that ΔH 0. The heat q may finally be obtained using the First Law, ΔU q + w. With ΔU 0, we have that q w 370 J.

9 9 6. Continued b.) Two moles of neon gas initially at 400ºC and.0 bar are expanded isothermally and irreversibly against a constant external pressure of 0.5 bar. Again, starting from the basic definition of work, we have w P ext dv. In this case, the process is not reversible. However, the process involves an expansion against constant external pressure, so we can pull the factor of P ext out of the integral. This yields V w P ext dv. Integrating leads to the expression w P ext ( V ). We again need to determine the initial and final volumes. Since the initial number of moles, pressure, and temperature are the same as in part (a), the initial volume is calculated from the ideal gas equation in the same way as before, n 1 P K ( mol) L bar mol 1 K 1 (.0 bar) 56.0L. Since the expansion will occur against the constant external pressure until the point that the final pressure P is equal to P ext, the final volume can be calculated as, V n P K ( mol) L bar mol 1 K 1 ( 0.5bar) V 447.7L. Finally, the work in Joules can be determined, w P ext ( V ) ( 0.5bar) ( 447.7L 56.0 L) 100 J 97.9 L bar 1L bar w 9790 J.

10 10 6 b.) Continued To determine the change in internal energy, we know that for an ideal gas, du C v dt, or in integrated form (given that C v is independent of temperature), ΔU C v ΔT. Since the process is isothermal, ΔT 0, and therefore we know that ΔU 0. To determine the change in enthalpy, we know that for an ideal gas, dh C p dt, or in integrated form (given that C p is independent of temperature), ΔH C p ΔT. Since the process is isothermal, ΔT 0, and therefore we know that ΔH 0. The heat q may then be obtained using the First Law, ΔU q + w. With ΔU 0, we have that q w 9790 J.

11 11 7.) (15 points) True/false, short answer, multiple choice. a.) True or False : A gas with a negative Joule-Thomson coefficient µ JT may be used as a refrigerant. U b.) True or False : Joule's experiment demonstrated that T V 0 for an ideal gas. c.) Short answer The heat capacity is a quantity that provides a measure of the ability of a substance to store energy. d.) Short answer A function that possesses an exact differential is one that is said to be path independent. e.) Multiple Choice: Circle the correct result for a reversible isobaric process in which an ideal gas undergoes an expansion from 1 L to 10 L. 1) w > 0, ΔU > 0, q > 0. ) w > 0, ΔU < 0, q < 0. 3) w < 0, ΔU > 0, q > 0. 4) w < 0, ΔU < 0, q < 0.