Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

------An ideal gaseous reaction (which is a hypothetical gaseous reaction that c

ID: 945995 • Letter: #

Question

------An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 30.0 atm and releases 75.0 kJ of heat. Before the reaction, the volume of the system was 8.40 L . After the reaction, the volume of the system was 2.20 L .(Calculate the total internal energy change, ?E, in kilojoules.)

Express your answer with the appropriate units.

------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?(Express your answer with the appropriate units)

Vacuum Gas

Explanation / Answer

1) The process is an isobaric process as it occurs at constant pressure.

Work done = - P(v2-V1) and from first law delta E = Q + W

Given P = 30 atm, V1 = 8.40 L , V2 = 2.2L and Q = -75kJ

Hnece work w = -30 (2.2-8.4) = 186 L.atm = 186 x 101.26 J

=18834.36 J

Delta E = q + w = -75000J + 18834.36 J = -56165.64 J = -56.165kJ

2) For the ideal gas , in isothermal irreversible compresion, w=q as delta E = 0

Thus in two step process w = w1 +w2

W1= -2(2.7 - 5.4 ) L atm = 5.4 L atm = 546.8J

W2 = -2.5 ( 2.16-2.7) = 1.35 L atm = 136.7 J

thus total work done = 683.5 J

heat liberated is -683.5 J

in single step

work done = -2.5 (2.16-5.4) = 8.1Latm = 820.2 J

heat liberated is -820.2 J

The difference in heat produced in the processes is 136.7 J