When 1.0 mole of liquid Br2 is converted to Br2 vapour at 25.0oC and 1.0 atm pressure, 7.3 Kcal of heat is absorbed, and 0.59 kcal of expansion work is done by the system. Calculate ΔU for this process
impatient much? reposting just wastes everyone's time.
dU = q + w
q = + 7.3 kcal.
w = -pdV = -p(V2-V1) = -59 kcal.
Note: If you use kcal for the units, obviously dU will be in kcal. Or do you want the units in joules. If so convert.
Plzz answer me
Well, well, well, we have quite the chemistry question here! ΔU, or change in internal energy, can be calculated using the equation:
ΔU = q - w
where q is the heat absorbed by the system and w is the work done by the system.
In this case, we have q = 7.3 Kcal and w = -0.59 Kcal (since the work done is negative if it is done by the system).
Plugging in the values, we get:
ΔU = 7.3 Kcal - (-0.59 Kcal)
ΔU = 7.3 Kcal + 0.59 Kcal
ΔU = 7.89 Kcal
So, the change in internal energy for this process is 7.89 Kcal.
Now, that's a lot of energy! I hope those bromine molecules are having a blast.
To calculate ΔU (change in internal energy) for this process, we need to consider the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat added to the system
W is the work done by the system
In this case, we are given that 7.3 kcal of heat is absorbed by the system. We also know that 0.59 kcal of expansion work is done by the system. We can use these values to calculate ΔU.
ΔU = Q - W
ΔU = 7.3 kcal - 0.59 kcal
ΔU = 6.71 kcal
Therefore, the change in internal energy (ΔU) for this process is 6.71 kcal.