Required information The state of an ideal gas with Cp= (5/2R is changed from P-

Question

Required information The state of an ideal gas with Cp= (5/2R is changed from P-1 bar and V-12 m3 to P2 = 4 bar and V-= 3 m3 Calculate Q, WAUt, and Ht by the following a mechanically reversible process. The values of R are given in the following table: Values of the Universal Gas C nstant 8.314 J.mol-1-K-1-8.314 m3.pa-mol-1-K-1 83.14 cm3-bar-mol-1-K-1-8314 cm3-kPa-mol-1-K-1 82.06 crn3.(atm)-mol-1 K-1-62, 356 cm".(torr)-mol-k-i 1.987 (cal)olK-11.986 (Btu)(Ib mole)(R) 0.7302 (ft)3 (atm) (lb mol)-i(R)--10.73 (ft)3(psia) (lb mol)-1(R)-1 1545 (ft) (lbr) (lb mol)-1(R)-1 R = = = Consider the process to be an adiabatic compression followed by cooling at constant pressure. (You must provide an answer before moving to the next part.) The heat required is -kJ The work done is kJ. The change in internal energy is The change in enthalnv is @kJ. k.l

Explanation / Answer

P1= 1 bar V1= 12 m3 and P2= 4 bar V2= 3 m3

P1V1= P2V2= 12 bar-m3

And for ideal gas PV = nRT

Therefore the whole process is isothermal process, i.e T1=T2

For isothermal process , heat required = Q= 0 KJ

And internal energy and enthalpy of ideal gas are function of temperature only and since temperature is constant so

Change in internal energy = 0 KJ

and enthaly = 0 KJ

For work calculation ,

Isothermal work = nRTln(V2/V1) ( n = number of moles )

so

W = P1V1ln(V2/V1) ( As  P1V1 = nRT)

= 1 * 105 * 12 ln( 3 /12)

= - 1663.55 KJ

Work is negative so work done by surrounding on system = 1663.55 KJ

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