8.535E-1 moles of an ideal diatomic gas undergoes a transition from a to c along

Question

8.535E-1 moles of an ideal diatomic gas undergoes a transition from a to c along the diagonal path in the figure. The temperature of the gas at point a is 1410 K. During the transition, what is the change in internal energy of the gas?

How much heat is added to the gas in going directly from a to c?

How much heat must be added to the gas if it goes from a to c along the indirect path abc?

(in case image is not showing up....A has a pressure of 5 kN/m^2 and a volume of 2 m^3, B has a pressure of 5 kN/m^2 and a volume of 4 m^3, and C has a pressure of 2 kN/m^2 and a volume of 4 m^3.

Explanation / Answer

part a:

temperature at C=pressure*volume/(number of moles*gas constant)

=2000*4/(8.314*0.8535)

=1127.4 K

so change in internal energy=number of moles*gas constant*temperature difference

=0.8535*8.314*(1127.4-1410)

=-2005.3 J

part b:

from A to C, pressure-volume equation:

(P-5000)/(V-2)=(2000-5000)/(4-2)=-1500

==>P=5000-1500*(V-2)=-1500*V+8000

work done=integration of P*dV

=integration of (-1500*V+8000)*dV from V=2 to 4

=-1500*(V^2/2)+8000*V

using the limits,

work done=-750*(4^2-2^2)+8000*(4-2)=7000 J

so heat added=work done+change in internal energy

=7000-2005.3

=4994.7 J

part c:

along the indirect path, change in internal eenrgy remains constant.

work done is zero for B to C as volume is constant

work done from A to B=pressure*change in volume

=5000*(4-2)=10000 J

so heat required=10000-2005.3

=7994.7 J

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