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 50.0 atm and releases 60.4 kJ of heat. Before the reaction, the volume of the system was 7.80 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules. Express your answer with the appropriate units. ΔE = -35.2 kJ SubmitHintsMy AnswersGive UpReview Part Correct Although the system absorbed some energy in the form of work, the significant release of heat caused the total change to be negative. State functions versus path functions The change in internal energy, ΔE, 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.

Part C 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 6.60 to 3.30 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.30 to 2.64 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 6.60 to 2.64 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.

please answer for Part C

Answer 1

1) Calculate w for the one step process. Take the change in volume 2.16L - 5.40L = -3.24L. Divide that value by 1000 to get -0.00324L. Then do (-0.00324L)(2.5atm*101325pascals)=(-0.00... Then take that value and divide it by 101.3J to get 8.10Liter-atmospheres. 8.10 is the value for work, w, for the one-step process.

2) Calculate w for the two-step process. Basically, repeat the same process with different values:
2.7L-5.4L=-2.7L. Take that value divided by 1000 to get -0.0027L. Take -0.0027L(101325pascals*2atm) to get -547.155J. Divide that by 101.3 to convert it to Liter-atmospheres, or L*atm, to get -5.4L*atm. Since this is a two-step process, the total work is the sum of the work of the first step and the second step. -5.4L*atm is the value of the work done in the first step. Now, we will calculate the work done in the second step and sum them to get the total work done in this two-step process. We take 2.16L-2.7L to get -0.54L. Divide that by 1000 to get -0.00054L. Take -0.00054L(2.5atm*101325pascals)=-136.789... Divide that by 101.3 to get -1.35L*atm. Now we sum together -5.4L*atm and -1.35L*atm to get -6.75L*atm, but we must flip the sign due to the negative sign in the formula -(P)(change in Volume). So we will get a final answer of a positive 6.75L*atm for the work done for the two-step process.

3) Now we calculate the difference between the q values in liter-atmospheres like the question asks. This is a rather simple step when given the formulas. q2 - q1 = w1 - w2. So, we just take our w1 value in the one-step process of 8.1L*atm and subtract the w2 value in the second-step process of 6.75L*atm to get 1.35L*atm. Lastly, we convert this value to Joules using 101.3J=1L*atm. So, 101.3J(1.35L*atm) = 136J, which is the final answer to this problem