Three moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2531 J of heat are added to the gas, and 1101 J of work are done on it. What is the final temperature of the gas?

delta U= 3/2nR(T final -T initial)

(2531J - 1101J) = 3/2(3.0mol)(8.31)(T final - 345K)

1430J = 37.395(T final - 345K)

1430J/ 37.395 = T final - 345K

38.24 + 345K = T final

383.2404K = T final

This answer is incorrect. Please explain to me where I went wrong.

In your calculation, you correctly applied the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system.

You correctly calculated the change in internal energy as (2531 J - 1101 J), which gives you 1430 J.

However, the mistake in your calculation is in applying the equation ΔU = (3/2)nRT, which is valid for an ideal monatomic gas only when there is no heat added or work done on the gas. In this case, heat is added to the gas and work is done on it, so this equation cannot be used to calculate the final temperature.

To determine the final temperature, you need to use the first law of thermodynamics in a different form, which takes into account the heat and work:

ΔU = Q - W = (3/2)nR(T final - T initial)

Rearranging the equation, we have:

T final = (ΔU + (3/2)nRT initial) / ((3/2)nR)

Substituting the values into the equation:

T final = (1430 J + (3/2)(3.0 mol)(8.31 J/mol·K)(345 K)) / ((3/2)(3.0 mol)(8.31 J/mol·K))

T final = (1430 J + 9096.855 J) / (37.395 J/K)

T final = 10527.855 J / 37.395 J/K

T final ≈ 281.32 K

So, the correct final temperature of the gas is approximately 281.32 K.