Question

1. (30pts) Derive expressions for a gas that obeys the Van der Waals equation of state of (P+a⁄v²)(v-b)=RT where v is specific volume and a and b are constants.

For an isothermal process derive expressions to calculate

1. change in enthalpy (h),
2. change in internal energy(u),
3. change in entropy (s),

Answer 1

the square of the density of gas, ox inversely poopor tional to the square of its specific volume "SP = a8 = 0/0² where the proportionality factor á has a definite numerical value for each 995: when these two coorections are intooduced, We 06tain the van der Waals Equation of state - Ans) (b) Change in Enthalpy ah= Cp dT hz-hi = Cp (T2-T) |hz-hi (T2 = T1) ©change in Entropy o Tds = dut Pdv Tds Ce dT + Pav ds=Cudz + P dv T (P+a) (0-6)=RT 29 ds=Cv dI + Rav of 4) * The coefficient 6 is intended to correct the volume Occupied by the gas molecules & is known as co-vorume T roig The term a is a correction that accounts for the molecular force of attraction, the term is called the force Of Cohesion Tasa dH - vap Tas = cp dT - vap ds=Cp dT _vdi T T ds= Cp dT - Rav T V Integrating S2-5,= Cven Tz + R ln V₂ ni 52-51=(pen Tz Ti Р Isothermal process (T) = Tz) @ Change in the internal Energy + R en P2 T2 =.5, du= CodT Uz-u, Cu (T2-T) S2:5, = R en V2 Vi Tuz-u, = 0 (Ans) (12= 1) 52-51 = R en P PLL ADS CSScanned with CamScanner