Question

What is the sign of AU as the system of ideal gas goes from point A to point B on the graph? (Hint the U is function of temperature only for a perfect gas) P (atm) 5 4 3 2 В 1 V (m3) 2 5 1 4 The states And B have the same internal energy, so delta U is zero The internal energy of the system increases, so delta U is positive The internal energy of the system decreases, so delta U is negative Delta U cannot be determined without knowing the process used (i.e., the path taken) to get form state A to state B

Answer 1

Question 1: The correct option is

The states A and B has the same internal energy, so delta U is zero.

Explanation: For an ideal gas, PV = RT, where all terms have their usual meaning. Here from the values in the plot, we see that PV value for A and B are same. Hence T is same for both cases. Since U is a function of T only, there will be no change of U for the process.

Question 2: The correct option is

Q provides energy output while W is zero.

Explanation: Since W = -P dV, here for going from A to D, dV is zero. So no work is done. Since PV = RT, Since P is decreasing at constant V, so T will also decrease and the extra heat will come out as output.

Question 3: The correct option is

W provides energy output, while Q provides energy input. They are equal in magnitude.

Explanation: Since U is a state function, and since the value of U is same at A and B. Since Q is lost in the process of going from A to D, Q will be gained for going from D to B. The Q supplied will perform work on the system which will be the output.