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p and T Dependence of Entropv Background Reading From the Math Case Study (p.17 of lecture outline), we have studied the p and T dependence
p and T Dependence of Entropv Background Reading From the "Math Case Study" (p.17 of lecture outline), we have studied the p and T dependence of entropy in a specific case already. Let us now discuss its use in ideal gas. First we have dS=(TS)pdT+(pS)Tdp The first term on the RHS is the T-dependence and the second term is the p-dependence. They have more useful equivalent expressions as mentioned in the lecture outline (p.17) leading to dS=TCpdT(TV)pdp For ideal gas, the second term on the RHS can be simplified using the ideal gas law (i.e., pV=nRT ), thus we have dS=TCpdTpnRdp. For a process with both p and T changes, the entropy change can be estimated by integrating this equation from an initial state (p1,T1) to a final state (p2,T2). That is, S=initialstalefinalstatedS=T1T2TCpdTp1p2pnRdp=T1T2TCpdT+nRln(p2p1) - End Reading Calculate the entropy change per mole when cadmium (Cd) vapor at 767C and 1atm pressure is heated to 1027C and compressed so that its final pressure is 6atm. Assume that the vapor behaves as an ideal monoatomic gas. (CV=12.5JK1mol1). Hint: Cp=CV+nR for an ideal gas . Ans: 10.25JK1mol1 Calculate (a) the heat absorbed, and (b) the entropy change when 0.5 mole of a perfect gas is allowed to expand at 300K from a volume of 1dm3 to a volume of 10dm3 against a constant pressure of 1atm. Assume no non pV-work is involved in the process. Hint: Use the equation of Question 16 in Assignment 1 for the isothermal expansion of an ideal gas. Ans: (a) 912J: (b) 9.57J/K
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