2) A massless wire is attached to the piston. When an external pressure of 2.00

Question

2)  A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 4.90 to 2.45 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.45 to 1.96 L .

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 4.90 to 1.96 L in one step.

If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

Explanation / Answer

For 2 step process:
Total work done = W1+ W2
W1= -P1 * (V1f- V1i)
     = - 2 * (2.45-4.9)
     = 4.9 atm L
W2= -P2 * (V2f- V2i)
     = - 2.5 * (1.96-2.45)
     = 1.225 atm L

W= W1+W2 = 4.9 + 1.225 = 6.125 atmL
Since 1 atmL equals 101.3 J
W = 6.125*101.3 = 620.46 J

For 1 step process:
work done = -P * (Vf- Vi)
     = - 2.5 * (1.96-4.9)
     = 7.35 atm L
Since 1 atmL equals 101.3 J
W = 7.35*101.3 = 744.6 J

Since Internal energy change is same for the cases and delta U = q+W
q(process 1) + W(process 1) = q(process 2) + W(process 2)
q(process 1) + 620.46 J = q(process 2) + 744.6 J
q(process 1) - q(process 2) = 124.1 J

Answer: 124.1 J

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