Question

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?

THE ANSWER IS NOT -101.3!!!!!

Answer 1

For the two-step process :

Change in volume in first step = 2.70 - 5.40 = - 2.70 L

Pressure = 2 atm

Work done = - pV = - 2 x (- 2.70)

                 = 5.40 L.atm

                = 547.155 J

Change in volume in second step = 2.16 - 2.70 = - 0.54 L

Pressure = 2.50 atm

Work done = - 2.50 x (- 0.54)

                 = 1.35 L.atm

                = 136.79 J

q1 = 547.155 + 136.79

q   = 683.9 J

For one step process :

change in volume = 2.16 - 5.40   = - 3.24 L

Pressure = 2.50 atm

Work done = - 2.50 x (- 3.24)

                   = 8.1 L.atm

q2              = 820.73 J

Difference between heat = q1 - q2

                                        = 683.9 - 820.73

                                        = 137 J