We have 1.62 mol of a gas at 600 K and 8 atm, which are subjected to an expansio

Question

We have 1.62 mol of a gas at 600 K and 8 atm, which are subjected to an expansion up to a volume of 50 L, performing for this a work of 3, 108.41 cal. By a different process and following a relation PV_y = cte, an expansion is carried out until reaching a temperature of 300 K and the work that is carried out has a value of 2, 430 cal; Then it is compressed in an isoenthalic manner, performing a work of -1, 560 cal, the recompressed until the work is equal to delta E and that delta H = 2, 430 cal considers that y = 1.40. Calculate the following A) All the variables of each state B) Heat, enthalpy, energy, total and partial works. Explain whether it is a cycle or not.

Explanation / Answer

Moles of gas, n= 1.62, T=600K, P=8atm, V= volume of gas= nRT/P= 1.62*0.0821*600/8=9.975 L

Assuming the expansion to be isothermal, the work = nRT*ln(V2/V1)=1.62*1.987Cal/mole.K*600*ln(50/9.975)=3113Cal

The expansion is isothermal. During isothermal expansion

P1V1= P2V2, P2= P1V1/V2= 8*9.975/50 =1.6 atm, pressure after isothermal expansion. Since the process is isothermal, deltaU( change in internal energy), deltaH( change in enthalpy), both of them will be equal to zero. From 1st law, deltaU= Q+W, Q= -W= -3113 Cal.

2. Second process, the gas is expansed under adiabatic conditins from the relation, PVy= constant

Work done = -n*R*(T2-T1)/(Y-1)=2414 Cal.

Since the process is adaibatic, Q=0 and deltaU= -2414 Cal.

deltaH= n*Cp*(T2-T1), since CP-CV= 1.987, CP/CV= 1.4, CP- CP/1.4= 1.987, CP= 6.95 Cal/mole.K

deltaH= -1.62*6.95*300=-3378 Cal.

For this process, T2/T1= (P2/P1) (Y-1)/Y,(300/600)= (P2/8)0.4/1.4

P2= 0.7 atm , the isothemral compression is given by -1.62*1.987*300*ln(P2/0.7)=-1560, P2= 3.5 atm, Pressure at the end of compression process. Change in internal energy for this process= 0,

the compresion is done at adiabatic conditions ( since deltaE= W).

deltaH = n*CP*(T2-T1) = 2430 , 1.62*6.95*(T2-300)= 2430, T2-300 = 516K

from P2/P1= (T2/T1) (Y-1)/Y, P2= 0.82 atm, W= -1.62*1.987*(516-300)/0.4= -1738 Cal,

deltaE= 1738 cal.

The process is not cyclicc, since final pressure condition of 1.6 atm is not reached.

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