Suppose a monatomic ideal gas is changed from state A to state D by one of the p

Question

Suppose a monatomic ideal gas is changed from state A to state D by one of the processes shown on the PV diagram below, in which interval on the vertical (P) axis corresponds to 8 atm, and the values of the lines on the horizontal (V) axis equal 5L, 10L, and 20L.

(a) Find the total work done on the gas if it follows the constant volume path A-B followed by the constant pressure path B-C-D.

________J

(b) Calculate the total change in internal energy of the gas during the entire process and the total heat flow into the gas.

change in internal energy __________J

Total heat flow into the gas ________J

Explanation / Answer

a)

work done from A to B is W_AB=0 ( at constant volume)

change in internal energy from A to B is,

dU=n*CV*dT

=n*3R/2*(T2-T1) (for monoatomic gas Cv=3R/2)

=3/2*(nRT2-nRT1)

=3/2*(P2*V2-P1*V1)

=3/2*(P2-P1)*V   (at constant volume V1=V2=V=5L)

=3/2*(8-16)*1.01*10^5*5*10^-3

dU=-6060 J ( A to B) (increases)

work done from B-C-D is W_BCD=P*dV ( at constant pressure)

dW=P*(V2-V1)

=8*1.01*10^5*(20-5)*10^-3

dW=12120 J

b)

now

heat supplied from B to D is,

dQ=n*Cp*dT

=n*3R/2*(T2-T1) (for monoatomic gas Cp=5R/2)

=5/2*(nRT2-nRT1)

=5/2*(P2*V2-P1*V1)

=5/2*(V2-V1)*P   (at constant pressure P1=P2=P=8atm)

=5/2*(20-5)*10^-3*8*1.01*10^5

dQ=30300 J

use the dQ=dU+dW (from B to D)

30300=dU+12120

dU=18180 J ( B to D)

total change in internal energy is,

dU=18180+6060=24240 J

and

Total heat flow into gas is,

dQ=6060+30300

dQ=36360 J

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