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 explan to me where I went wrong.

if work was done on it, wouldn't the internal energy go up, and therefore the work ADDED to the heat added? Maybe I don't understand your statement.

since work is done on the gas, the value of work has to be negative. so you would do 2531 + 1101 instead

You are correct that if work is done on the system, the internal energy would increase. In this case, the work done on the gas is positive (+1101 J), indicating that it gained energy from the surroundings. However, the equation you used to calculate the change in internal energy (ΔU) is incorrect.

The correct equation for ΔU is:

ΔU = Q - W

where ΔU is the change in internal energy of the gas, Q is the heat added to the gas, and W is the work done on the gas.

Using the given values:

ΔU = 2531 J - 1101 J
= 1430 J

Now, to find the final temperature (T_final), we can use the equation:

ΔU = (3/2)nR(T_final - T_initial)

Substituting the values:

1430 J = (3/2)(3.0 mol)(8.31 J/mol·K)(T_final - 345 K)

Simplifying the equation:

1430 J = 37.395 J/K·mol (T_final - 345 K)

Dividing both sides of the equation by 37.395 J/K·mol:

38.24 K = T_final - 345 K

Adding 345 K to both sides of the equation:

T_final = 383.24 K

Therefore, the correct final temperature of the gas is 383.24 K.

You are correct in your understanding that if work is done on the gas, the internal energy of the gas increases and the work done should be added to the heat added. However, in your calculation, it seems like you subtracted the work done from the heat added, which is why you obtained an incorrect answer.

The correct equation to use is the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W).

ΔU = Q - W

In this case, the heat added to the gas is 2531 J, and the work done on the gas is 1101 J. So the equation becomes:

ΔU = 2531 J - 1101 J

Now, since the gas is an ideal monatomic gas, we can use the equation for the change in internal energy of an ideal monatomic gas:

ΔU = (3/2)nR(T final - T initial)

Substituting the given values, the equation becomes:

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

1430 J = 37.395 J/K·mol (T final - 345 K)

Now, dividing both sides of the equation by 37.395 J/K·mol:

1430 J / 37.395 J/K·mol = T final - 345 K

38.24 K = T final - 345 K

Adding 345 K to both sides of the equation:

38.24 K + 345 K = T final

383.24 K = T final

So the correct final temperature of the gas is 383.24 K.