ACS Physical Chemistry Thermodynamics Exam Study Guide, Exams of Physical Chemistry

Download ACS Physical Chemistry Thermodynamics Exam Study Guide and more Exams Physical Chemistry in PDF only on Docsity! ACS Physical Chemistry Laws of Thermodynamics and State Functions. Mathematical Relationships in Thermodynamics Exam Study Guide Boyle's law p and v are inversely proportional for a gas (by holding T and n constant) definition of reduced parameters in terms of critical parameters for N2 Tr= 126 K/ 126 K=1 and Pr= 1atm/39 atm=0.026 Previous Play Next Rewind 10 seconds Move forward 10 seconds Unmute 0:04 / 0:15 Full screen Brainpower Read More the derivative of the enthalpy with respect to P at constant T, (dH/dP)T, for an ideal gas is 0 the Joule-Tompson coefficient will predict whether A real gas heats or cools on pressure change the sign of the Joule-Tompson coefficient can be predicted from the equation of state of the real gas in the van der Waals equation of state, P=(nRT/(V-nb))-((n^2a)/V^2)), the terms, nb, will increase as the molecular diameter increases under what condition is H2 in the state corresponding to N2 at 126 K and 1 atm? 33 K, .33 atm the isothermal compressibility kT = -(1/V)(dV/dP)T, for the hard sphere equation of state P(V-nb) = nRT is given by (RT/P^2)/(RT/P +b) definition of reduce temperature Tr=T/Tc reduced pressure Pt=P/Pc definition of the corresponding states it must have the same reduced temperature and pressure the valve between the 2.00 L bulb, in which the gas pressure is 1.00 atm, and the 3.00 L bulb, in which the gas pressure is 1.50 atm, is opened. What is the final pressure in the two bulbs, the temperature being constant and the same in both bulbs? After the valve is opened, the total volume is (2.00 + 3.00) L + 5.00 L and the total number of moles is the same of moles initially in separate bulbs. Since both, the volume and number of moles add, 1.00 atm2.00 L + 1.50 atm3.00 L = P*5.00 L so P=1.30 atm ideal gas equation of state at constant temperature the number of moles is directly proportional to the product of pressure and volume, PV what kind of isotherms graph is experimentally obsessed near the critical temperature of the real gas? negative wave like exponential graph This shows P - V behavior of real gas near the critical point. The critical point itself occurs when (∂P/∂V)T = 0 and (∂^2P/∂V^2)T=0 as pressure and temperature are increased to the critical point... -Δ(vaporization)H goes to 0 -the density of the fas approaches the same value as that of the liquid -the index of refraction of the gas approaches the same value as that of the liquid the van der Waals equation of state (P+(n^2a/V^2))(V-nb)=nRT contains a term representing a "molecular size". The approximate magnitude of this term is 1 cm^3/mol terms in real gas equations of state and their approximate magnitude when the external pressure is zero (as is the case when expansion takes place into a vacuum) then no work is done in the process expansion the system does work on the surroundings, so w<0 and U will decrease In an adiabatic expansion of an ideal gas, what is always true for w done on the surroundings and ΔU? The work done by the gas on the surroundings is equal to the decrease in the internal energy of the gas T will go down Because the expansion is adiabatic, we know that q = 0; then because ΔU = q + w, it follows that ΔU = wad. In an expansion-the system does work on the surroundings, so w<0 and U will decrease. If an adiabatic expansion of an ideal gas occurred into a vacuum, is work done on the gas by the surroundings? No work is done on the gas by the surroundings Pext=0 If each CO molecule in a carbon monoxide crystal has equal probability of being situated on a lattice site with one of the two orientation CO or OC, the initial value of S at 0 K will be nearest 5.76 J/Kmol q and ΔU for an adiabatic process where the system decreases in volume? q=0 ΔU>0 When a transformation occurs spontaneously at constant T and P, the signs of ΔG for the system and ΔS for the universe must be ΔGsystem: negative ΔSuniverse: positive The 2nd law states that ΔSuniverse must be positive for a spontaneous process calculation of third law entropy from heat capacity data S=S₀+∫Cp/T dT from initial temperature to final temperature As the temperature approaches absolute zero, ΔG for any chemical reaction approaches ΔH ΔG=ΔH-TΔS. As T goes to 0, the second term goes to 0 thus ΔG goes to the value of the first term, ΔH ΔU for the reaction A(s) + 2B(g) --> 2C(s) + D(g) was measured at temperature T in a constant volume calorimeter. ΔH for the reaction is approximately.... ΔU - RT H = U + nRT  n for the gas (g) = product gas moles-reaction gas moles = 1 - 2 = -1 therefore , H = U - R*T an ideal gas undergoes irreversible isothermal expansion so.... ΔS= -ΔG|T the efficiency of a process with q1 occurring at Th and q2 occurring at Tc<TH can be calculated from |w|/q1 a measure of the maximum non-PV work that can be performed by a process occurring at constant T and P is given by ΔG the entropy change for a liquid heated for T1 to T2 can be calculated from the area under the curve obtained by plotting Cpas the ordinate and and let as the abscissa in a cyclic process involving two steps, the first law of the thermodynamics requires that |q1 + q2| = |w1 + |w2| The third law of thermodynamics can be combined with experimental data to provide an absolute value for entropy total differential dH gives gives the total change in H arising from change in both U and Pv. The change in U is represented by differential dU and the change in PV is represented by the differential d(PV)Since both P and V are independent variables d(PV) is equivalent to VdP the total differential for H is dU + VdP + PdV if the effect of pressure on a reaction involving only pure solids is taken into account, then ΔG(T) = ΔGo(T) + ΔV(P-Po) The molar heat capacity of diamond is adequately given by the equation Cp/(Jk-1mol-1) = 3.02x10^-7 T^3. How much heat does it take to raise the temperature of one mole of diamond from 100 k to 300 k? 1/4 (3.02 x 10^-7)(300^4-100^4) Given the heat capacity information above, what (Cp/(Jk-1mol-1) = 3.02x10^-7 T^3) is the absolute entropy of diamond at 300 K? 1/3 (3.02 x 10^-7)(300)^3