Question

Cycles in a heat engine The basic idea of a heat engine is to add heat to a gas, and have the gas perform work on its environment via an expansion, which in turn is used to power motion. The gas is then restored to its original temperature and the cycle begins again. You can assume that the same amount of gas remains in the engine during the cycle. A convenient way to think about these cycles is a pressure vs. volume plot.

(a) Suppose a gas is follows the rectangular path shown in the P‐V plot.

P 4P. b с Po a d →V Ve 4V.

For each of the parts of the cycle, use the first law of thermodynamics conservation of energy), Q = W + ΔU, where Q is the heat added to the gas, W is the work done by the gas, and ΔU is the change in the internal energy ΔU=(5/2)nRΔT  of the gas. Assume we are using a diatomic gas like N2, and treat it as an ideal gas so PV = nRT. Fill in the table with the sign (+, 0, or ‐) of each of these quantities:

(b) If the temperature of the gas at point a is T0, what is the temperature at points b, c, and d?

For the following parts, express your answers in terms of only P0 and V0.

(c) What is the internal energy of the gas at points a, b, c, and d?

(d) How much heat is added to the gas during the heating phase (from a to b)?

(e) How much work is done by the gas during the expansion phase (from b to c)?

(f) How much heat is added to the gas during the expansion phase (from b to c)?

(g) How much work is done by the gas during the compression phase (from d to a)? What is the net work done by the gas during the entire cycle?

(h) What is the ratio of the total heat added to the gas (during the heating and expansion phases) to the net work done by the gas?

Answer 1

4 a d the gas. + From first law of thermodynamics- the W + AU where, Q = heat supplied to the gas. Work done by Po Change in internal energy 4vo of the gas, For process a to b Volume constant pus NRT P = MR here n, R and v are constant • P LT i To Ta Po To where, Ta= Temperature at a. To = Temperature at ba = 4Ta Now, Wtou = P(Vb-Va) + 5 MR (To-Ta) SnR ( 4ta - Ta) - [Vo va 15 nRTA W=0 Q= Vb To 15 nRTA 2 ve Tc :: For process a to b- a + W o vo Tь AVO TC 3 Tc = 456 AU = + Q = WUDU process to to C P= constant VLT '' and su= 5nR[Tc - To Dy=+ W 24P| Ve-V.) A w=t QE+

___ For process c. to da Volume constant : W = Pl Vd-Vc) o [ Vd = vc] PV = nRT V = constant Pc Tc po Td 400 To => Pd = Ta V Te = 4td Ta = Tc DUE MR (Td-Tc) TdLT DU > from First law of thermodynamics- Q = (-)+(0) Du tw DU SNR (Ta-Td) Tac Td DU > - . For process d to att p = constant W = P. (va-va) Polvo 4V.) 3 Po. Vo VXT > td Va > Avo W -- Td Ta - Td Vo > Ta = 4T Ta = Ta 4 Ta Dut w (-) + (-) Q =

a to b + + Ibtoc + + + c to d - ulo Id to a JA VA otlot TA IA - if Ta= To For process ab (1= constant) Pv= MRT Pa = Ta = To Pb Tb Po To 410 Tb Tb= 4To process boc (P= constant) Pv=ART VaT Vc To Tc Vb To 4To FIFA © Internal energy 돌까 NRT For va = orta 7 PV = nRT =) Poo Vo=nRTa Ta = Porvo NR i va = DR Ź AVO Tc 4To Vo NR 0 Te- 16 To Va – Polo Process cod (V= constant) PV = nRT For point b- Tc = 1670 Ta Pd Th= 4 po x V. Pc Td > 400 16 To Po Ta Td = 4 To -MR Ub= ORTO 4 Povo * Ub={nR NR Ub - 10 Povo.

for point co Tc = (4P.). (Avo) NR 16 Porvo пR Uca NRTC > Uc= $na 16 Povo NR Uc = 40 po. No For point d Td = (Pax(4.v.) = 4 Povo na MR .: Ud &nota. Vd = & na [ 4 Povo HR Ud=10 Povo @ heating phase (a tib) :- Autw &Q (U6-Ua) + 0 [W= WF O bebure Volume is constant] => p = (to Povo – Ş povo) 7.5 Povo Q = is Povo Q Q
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