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.
Delta U would be 2531J + 1101J instead of 2531J - 1101J, because the system is doing work instead of having work done to it.
(2531J + 1101J)= 3632J
3632J = 3/2(3.0mol)(8.31)(T final - 345K)
3632J = 37.395(T final - 345K)
3632J/ 37.395 = T final - 345K
97.12 + 345K = T final
442.125K = T final
The mistake lies in incorrectly calculating the value of the gas constant (R). The gas constant (R) for an ideal monatomic gas should be 8.314 J/(mol*K), not 8.31 as used in the calculation. This difference in the value of R causes the final result to be slightly incorrect.
To get the correct answer, we need to use the correct value of R in the equation:
delta U = (3/2)nR(Tfinal - Tinitial)
First, let's calculate the correct value of R:
R = 8.314 J/(mol*K)
Now, let's substitute the values into the equation:
delta U = (3/2)(3 mol)(8.314 J/(mol*K))(Tfinal - 345 K)
(2531 J - 1101 J) = (3/2)(3 mol)(8.314 J/(mol*K))(Tfinal - 345 K)
1430 J = (3/2)(8.314 J/(mol*K))(Tfinal - 345 K)
1430 J = 12.471 J/(mol*K)(Tfinal - 345 K)
Now, divide both sides of the equation by 12.471 J/(mol*K):
1430 J / 12.471 J/(mol*K) = Tfinal - 345 K
114.852 mol*K = Tfinal - 345 K
Add 345 K to both sides of the equation:
114.852 mol*K + 345 K = Tfinal
459.85 K = Tfinal
Therefore, the correct final temperature of the gas is approximately 459.85 K.