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 55.7 kJ of heat. Before the reaction, the volume of the system was 8.00 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.

Answer 1

The process occurs at constant pressure of 30.0 atm and the initial and final volumes of the gaseous system are 8.00 L and 2.80 L. Assume the compression takes place under constant external pressure of 30.0 atm and find out the work done is

W = -∫P.dV where P = external pressure of the system and dV = change in volume of the system. Since the compression takes place at constant pressure, we can write

W = -P_ext.ΔV where ΔV = V₂ – V₁ is the change in volume of the system. Given V₁ = 8.00 L and V₂ = 2.80 L, we have,

W = -(30.0 atm)*(2.80 – 8.00) L = -(30.0 atm)*(-5.20 L) = 156.0 L-atm = (156 L-atm)*(101.33 J/1 L-atm) [1 L-atm = 101.33 J] = 15807.48 J = 15.80748 kJ (I will keep a few guard digits extra).

The system releases 55.7 kJ heat. Since heat is released by the system, using the sign convention (heat transferred by the system is negative), we have Q = -55.7 kJ.

The change in internal energy of the system is given as ΔE = Q – W = Q + (-W) = (-55.7 kJ) + (-15.80748 kJ) = -71.50748 kJ ≈ -71.51 kJ (ans).

You may obtain an answer of -39.89 kJ depending on the sign convention you used. I used the following sign convention:

Heat released, Q is negative

Work done, W is positive.

If you use the convention, heat released, Q is negative and work done, W is negative, then you will get the value of -39.89 kJ. Please check the proper sign convention with you TA or instructor.