Problem Set 4 Solutions

Transcription

1 Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal reversible expansion of an ideal gas, the work is given by w T ln V & % ( The initial and final volumes are not given However, for an isothermal process, from the ideal gas equation we have P 1 P V, or Substituting, the work can be calculated as V P 1 P w T ln P & 1 % ( P 98 K ( mol) 8314 J mol 1 K 1 w 650 J & ( 00 atm 1atm ln% For an ideal gas, the differential of the internal energy is given by du C v dt and the differential of the enthalpy is given by dh C p dt Since the process is isothermal, dt 0, and so du 0 and dh 0 Therefore, Finally, from the First Law of thermodynamics, Summary of results: ΔU 0 and ΔH 0 ΔU q + w Since q w 650 J ΔU 0, q w ΔU ΔH 650 J 650 J 0 J 0 J

2 A balloon filled with 0505 mol of gas contracts reversibly from 10 L to 010 L at a constant temperature of 50 C It simultaneously loses 170 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 Substituting, the work can be calculated as w T ln V & % ( w T ln V & % ( 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 170 J, so the First Law yields 7815K ( 0505 mol) 8314 J mol 1 K 1 w 690 J 010 L ln% 10 L & ( ΔU q + w 170 J J ΔU 140 J Finally, to determine the enthalpy change ΔH the definition of enthalpy may be used, H U + PV ΔH ΔU + Δ( PV) In order to determine the change ideal gas equation may be employed to determine the initial and final pressures, and Δ( PV) P V P 1, the volumes are given but the pressures are not So, the P 1 T 7815K ( 0505mol) 00805L atmmol 1 K 1 P atm ( 10 L) P T V 7815K ( 0505mol) 00805L atmmol 1 K 1 P 1155atm ( 01L)

3 Continued 3 Now the change Δ( PV) may be determined, Δ( PV) P V P 1 ( 1155atm) ( 01L) ( 1155atm) ( 10 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( T ) ΔT Δ PV Δ PV Since the temperature is constant, therefore given by ΔT 0 and as shown above, ΔH ΔU + Δ PV ΔU ΔH 140 J Δ( PV) ΔT 0 The enthalpy change is Summary of results: q w ΔU ΔH 170 J 690 J 140 J 140 J 3 The temperature dependence of the molar constant pressure heat capacity of a real gas may be represented by the function C p,m α + β T + γ T where α, β, and γ are constants For N, these constants have the values α 6984 J K 1 mol 1, β J K mol 1, and γ J K 3 mol 1 Determine the amount of heat required to raise the temperature of 1 mol of N gas from 300 to 1000 K at constant pressure In this problem, we know that the heat at constant pressure is the enthalpy, dq p dh The exact differential of enthalpy is dh C p dt + H & % ( dp P T

4 3 Continued 4 At constant pressure, dp 0, so the exact differential of enthalpy reduces to On a molar basis, this equation becomes dh C p dt dh m C p,m dt Integrating, H,m T dh m C p,m dt H 1,m Substituting the values of the parameters, we have Thus, at constant pressure, T dt ΔH m α + β T + γ T ΔH m α ( T ) + β T ΔH m α ( T ) + β T + γ 3 T 3 3 ( ) + γ 3 T 3 3 ( ) ( 1000 K 300 K) + 1 ( J mol 1 K ) ( * 1000 K ) ( J mol 1 K 3 ) ( * 1000 K ΔH m 6984 J mol 1 K J/mol J/mol 110 J/mol ΔH m 1468 J/mol ( 300 K) + -, 3 ( 300 K) 3 ) + -, q p,m ΔH m 1468 J/mol 4 Estimate the final temperature of one mole of gas at 000 atm and 190 C as it is forced through a porous plug to a final pressure of 095 atm The Joule-Thomson coefficient of the gas is 0150 K/atm The Joule-Thomson coefficient µ JT is defined as µ JT T & % ( P Since the problem requests an estimate of the final temperature, we can approximate the derivative in the following way, µ JT T & % ( P H H ΔT ΔP Note that in the equation above we have approximated infinitesimal changes by finite changes

5 4 Continued 5 We can solve the expression for the temperature difference ΔT, µ JT ΔT ΔP or ΔT µ JT ΔP Substituting the numerical values, ΔT µ JT P P 1 ( 095atm 000 atm) 0150 K/atm ΔT 986 K The final temperature then may be determined, ΔT T 986 K or T 986 K T 916 K 986 K T 69 K or 109! C So for a pressure drop of nearly 00 atm, the temperature of this particular gas drops by nearly 30 degrees 5 Determine the final temperature of 01 mol of an ideal monatomic gas that performs 75 J of work adiabatically if the initial temperature is 35 C For an adiabatic process, q0, so from the First Law, ΔU q + w ΔU 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 C v dt (for an ideal gas) U T du C v dt T1 U 1 ΔU C v T1 T ΔU C v 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 dt ΔU w C v ΔT

6 5 Continued 6 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 Substituting, C v 3 ΔU w C v ΔT 3 ΔT We can now solve this equation for the temperature difference, ΔT w 3 ( 75J) ( 8314 J/molK) 3 01 mol ΔT 499 K Note that the work is negative because the system performs work on the surroundings The initial temperature is 35ºC or K Solving for the final temperature, we have ΔT T 499 K or T 499 K T 50815K 499 K T K or 1857! C 6 A tank contains 0 L of compressed nitrogen at 10 atm and 5 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, w ΔU ΔU q + w With q 0, we have For an ideal gas, the differential of the internal energy is given by U T du C v dt U 1 Assuming that C v is independent of temperature, the integration yields w ΔU C v ( T ) du C v dt Integrating, 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 7 For a reversible adiabatic process, we derived in class that Using properties of logarithms, we have C v ln T % ln V % & & ln T % C v & ln V % & Taking the exponential of both sides yields T % C v & V % &, or T % C v & V % 1 V & For an ideal gas, give V T P, so the ratio of volumes on the right hand side of the equation can be rewritten to T % C v & P % T P 1 & Rearranging to get the temperatures on the left, T % C v T % & & C T % v + & T P P 1 % & P % P 1 & P % P 1 & C v + Solving for the final temperature, P % T T 1 & P 1 P % T T 1 & P 1 C v +, C p

8 6 Continued 8 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 C p,m 915 J mol 1 K 1 Therefore, the exponent is C p nc p,m R C p,m 8314 J mol 1 K 1 915J mol 1 K 1 C p The final temperature is P % T T 1 P 1 & 9815 K T 1545 K atm % 10 atm& Now, the work and internal energy change can be calculated using the relation w ΔU C v T nc v,m ( T ) Since C p C v +, in terms of molar quantities, we have C p,m C v,m + R We can solve for to get C v,m C p,m R 915J mol 1 K J mol 1 K 1 C v,m 0811J mol 1 K 1 The moles of nitrogen gas can be calculated from the initial state, C v,m n P 1 R ( 10 atm) ( 0 L) ( 00805L atmmol 1 K 1 ) 9815K n 8176 mol Substituting, w ΔU nc v,m ( T ) ( 8176 mol) 0811J mol 1 K 1 w ΔU 4440 J ( 1545 K 9815K)