Question

Part B: An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 45.0 atm and releases 70.4 kJ of heat. Before the reaction, the volume of the system was 9.00 L . After the reaction, the volume of the system was 3.00 L .

Calculate the total internal energy change, ΔU, in kilojoules.

Part C:

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 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 5.40 to 2.70 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 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 5.40 to 2.16 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?

Answer 1

Part B)

work = - p dV

W = 45 x (9.00 - 3.00 )

      = 270 L.atm

     = 27.35 kJ

Q = 70.4 kJ

dU = dQ +   d W

    = - 70.4 + 27.35

dU = - 43.0 kJ

internal energy = - 43.1 kJ

Part B)

When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.40 to 2.70L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16L .

Step 1 :

w1 = - p (V2 - V1)

    = 2.00 (2.70 - 5.40)

   = 5.4 L.atm

   = 547.02 J

w2 = 2.50 x (2.16 - 2.70)

     = 1.35 L.atm

     = 136.76 J

w = w1 + w2 = 683.78 J

step 2 :

w3 = 2.50 x (2.16 - 5.40 )

   = 8.1 L.atm

   = 820.53 J

change in work = w - w3

                         = 683.78 - 820.53

change in work = - 137.J