Hello, I am struggling with finding the answers for B and C. May you please help

Question

Hello,

I am struggling with finding the answers for B and C.

May you please help me

Thank you so much

An ideal gaseous reaction occurs at a constant pressure of 40.0 atmand releases 70.5kJ of heat. Before the reaction, the volume of the system was 9.00L. After the reaction, the volume of the system was 2.60L. Calculate the total internal energy change, Delta U, in kilojoules. Express your answer with the appropriate units. The change in internal energy, Delta U, is a state function because it depends only on the initial and final states of the system, and not on the path of change. In contrast, q and w are path functions because they depend on the path of change and not just the initial and final states of the system. An ideal gas is 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 is increased to 2.50 atm, the gas further compresses from 3.30 to 2.64L. 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 6.60 to 2.64L 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

Part B

Enthalpy change = Change in internal Energy+ Work done

and Wok at constant Pressure= Pressure*Change in Volume

Thus

-70500 = Changein int. En. + ((40*10^5 N/m^-2)((2.6*10^-3)-(9*10^-3) m^3)

Thus Change in Internal Energy = -44.9kJ

Part C

Since the final state is the same in both the processes therefore,the change in the internal energies are same in both the processes.Thus for the two step process from first law we have:

?U = q1

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