Problem 1: A system of ideal gas within a piston-cylinder assembly at (po, Ti) i

Question

Problem 1: A system of ideal gas within a piston-cylinder assembly at (po, Ti) is surrounded by the environment at (Po, To) (see Fig. la). You are asked to find the maximum work that can be extracted from this system-environment pair, i.e., when the system attains a dead state of (Po, To) (Fig. ld). To achieve that, we will take the system through a process involving adiabatic expansion from (po,Ti) to (Pow, To), and isothermal contraction from(to (po. To). a. T) At any intermediate state, when the system is at (p, T), mass has to be added to the pan such that a force of (po-P)A is exerted on the piston to keep it in equilibrium. Answer the following questions: 1. Show that work done on the system-surrounding pair is (A-P)dV, where Vi and ½ are the volumes at the initial and final states respectively 2. Assuming that the system consists of n kilomoles, find the initial volume V 3. Find the pressure Piow and volume Vhigh at the end of the adiabatic expansion process. 4. Find the work done by the system-surrounding pair in the adiabatic expansion process from (po,T) to (P.To) from (prov ,T,) to (Po,%). (Po, T) to the final "dead state (po,To) 5. Find the work done by the system-surrounding pair in the isothermal contraction process 6. Find the overall work extracted from the system when it is taken from an initial state . Initial state (pi,T . Intermediate state (Plow, To) Final state (Po,To Isotherm at T Isotherm at To Adiabat at T Adiabat at To Volume Figure 1: (a) System (enclosed within the piston and the fixed end wall of the cylinder) is at initial state (po, T. (b) The mass is increased so that a force of (p p)A is exerted on the piston. (c) The piston reaches the extreme position at which T-T) and p-Plow. (d) The mass is changed such that the system is compressed isothermally at To from (po ,%) to (Por) (e) p-V diagram for the process from (a) to (d). Though explicit pressure values are not given, it can be inferred from the shape of isotherms that the pressure axis (y-axis) is plotted on a log-scale. POST QUESTION 19 questions remaining this

Explanation / Answer

given step 1 : adiabatic expansion from po, Ti to plow, To

step 2 : isothermal contraction from plow, To to po, To

1. work done in adiabatic expansion = Wa

now, for a samll volume change dV

dWa = podV ( as external pressure always remains po, and work is done against external pressure)

for isothermal compression

dWi = -pdV [ as work is done on the gas and internal pressure at some time t is p]

hence net work done by the gas is

dW = dWa + dWi = integrate[(po - p)dV] from Vi to Vf ( where Vi and Vf are initial and final volumes respectively)

2. for n kilomoles of gas in the system, let initial volume be Vi

then from ideal gas equation

poVi = 1000nRTi

Vi = 1000nRTi/po [ where R is universal gas constant]

3. at the end of the adiabatic process

plow*Vhigh^(gamma) = po*Vi^(gamma) [ as in adiabatic process, PV^gamma is constant]

plow = po*(1000nRTi/po*Vhigh)^gamma

now from ideal gas equation

plow*Vhigh = 1000nRTo

po*(1000nRTi/po)^gamma*Vhigh^(1 - gamma) = 1000nRTo

Vhigh = [1000nRTo(po/1000nRTi)^gamma/po]^(1/(1-gamma))

hence

plow = po*(1000nRTi(po/1000nRTo(po/1000nRTi)^gamma)^(1/(1-gamma))/po*)^gamma

4. work done in adiabatic expansion = Wa

Wa = po(dV) = po(Vhigh - Vi)

where Vhigh = [1000nRTo(po/1000nRTi)^gamma/po]^(1/(1-gamma))

Vi = 1000nRTi/po

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