When a system is taken from state a to state b in the figure ( Figure 1 ) along

Question

When a system is taken from state a to state b in the figure (Figure 1) along the path acb, 89.0J of heat flows into the system and 63.0J of work is done by the system.

A.) How much heat flows into the system along path adb if the work done on the system is 12.0J?

B.) hen the system is returned from b to a along the curved path, the absolute value of the work done by the system is 36.0J . How much heat does the system liberate?

C.) If the internal energy is zero in state a and 6.0J in state d, find the heat absorbed in the processes ad.

D.) Find the heat absorbed in the processes db.

Explanation / Answer

A)

Along Path acb :

Qacb = 89 J

Wacb = 63 J

Uacb = ?

Using first law of thermodynamics

Qacb = Wacb + Uacb

89 = 63 + Uacb

Uacb = 26 J

for Path adb :

Uadb = Uacb = 26 J

Wadb = - 12 J

using first law of thermodyanamics

Qadb = Wadb + Uadb

Qadb = - 12 + 26

Qadb = 14 J

B)

Uba = - Uadb = - Uacb = - 26 J

Wba = 36 J

Qba = ?

using first law of thermodyanamics

Qba = Wba + Uba

Qba = 36 + (-26)

Qba = 10 J

C)

Uad = 6 J

Wad = Wadb = 26 J

Qad = ?

using first law of thermodyanamics

Qad = Wad + Uad

Qad = 26 + 6

Qad = 32 J

d)

Uad = 6 J

Uadb = 26 J

Udb = ?

Udb = Uadb - Uad = 26 - 6 = 20 J

Wdb = 0

hence Qdb = Udb + Wdb

Qdb = 20 + 0

Qdb = 20 J

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