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 6.40 to 3.20 LL . When the external pressure is increased to 2.50 atmatm, the gas further compresses from 3.20 to 2.56 LL .

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 6.40 to 2.56 LL 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?

Express your answer with the appropriate units.

Answer 1

According to first law of thermodynamics

The change in internal energy U = Q - W

Where Q is the heat supplied, and W is the work done.

In process 1: the compression is done in two stages.

                   In stage 1 the external pressure is 2 atm and the volume is compressed from 6.4L to 3.2L

      Therefore U₁ = Q₁ - W₁ = Q₁ - P_ext ( V₂ -V₁)

                            = Q₁ - 2 ( 3.2 - 6.4)

                           = Q₁ + 6.4

Similarly in stage 2 the external pressure is 2.5 atm and the volume is compressed from 3.2 L to 2.56L

Therefore U₂ = Q₂ - W₂ = Q₂ - P_ext ( V₂ - V₁)

                      = Q₂ - 2.5 ( 2.56 - 3.2)

                     = Q₂ + 1.6

Therefore total change in internal energy U₃ = U₁ + U₂

                                                                    = (Q₁ + Q₂) + (6.4 + 1.6)

                                                                U₃    = (Q₁+ Q₂) + 8 ----------------- (1)

In the second process, the compression is effected in one single step . However the final temperature of both process is the same. (given)

Therefore the change in internal energy of the single step process 2 is equal to the total change in internal energy of the two step process 1

Here the external pressure is 2 atm and the volume is compressed from 6.4 L to 2.56 L

      Therefore U₃ = Q₃ - W₃ = Q₃ - ( 2.56 - 6.4)

                            U₃ = Q₃ + 7.68 ---------------------- (2)

From equations (1) and (2)

                Q₁+ Q₂ + 8 = Q₃ + 7.68

               (Q₁ + Q₂ ) - Q₃ = 7.68 - 8

                                     = - 0.32 L-atm

That is q (process 1) - q (process 2) = - 0.32 L-atm

                                                     = - 0.32 x 10^-3 m³ x 1.01325 x 10⁵ Pa

   [ since 1 atm = 1.01325 x 10⁵ Pa and 1 L = 10^-3 m³ ]

                                                    = - 0.32 x 10^-3 m³ x 1.01325 x 10⁵ N/m² [ since 1 Pa = 1 N/m² ]

                                                    = - 0.32 x 101.325 N-m

                                                   = -32.4 N-m

                                                  = - 32.4 J         [since 1 N-m = 1 J ]

Thus the difference between q for the two step process and q for the one step process is -32.4 J.

Thus the heat involved in the two step process is less by 32.4 J.

                                                 

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