If 244 KJ heat is released and 35 KJ WORK done on the system.calculte (a)change internal energy and change in enthalpy of system? (b) change in internal energy and change in enthalpy of the suround ?
To calculate the change in internal energy (ΔU) and change in enthalpy (ΔH) of the system and surroundings, we need to use 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.
Given information:
- Heat released by the system (Q) = -244 kJ (negative sign because heat is released)
- Work done on the system (W) = 35 kJ
(a) Change in internal energy and change in enthalpy of the system:
To calculate ΔU, we use the equation:
ΔU = Q - W
Substituting the given values:
ΔU = -244 kJ - 35 kJ
ΔU = -279 kJ
So, the change in internal energy of the system (ΔU) is -279 kJ.
For an ideal gas, the change in enthalpy (ΔH) is given by:
ΔH = ΔU + PΔV
If the volume remains constant (no change in volume), then ΔV = 0 and PΔV = 0. Therefore, the change in enthalpy (ΔH) of the system would be equal to the change in internal energy (ΔU).
ΔH = ΔU = -279 kJ
(b) Change in internal energy and change in enthalpy of the surroundings:
Since energy is transferred from the system to the surroundings, the change in internal energy of the surroundings (ΔU_sur) would be equal in magnitude but opposite in sign to the change in internal energy of the system:
ΔU_sur = -ΔU = 279 kJ
Similarly, the change in enthalpy of the surroundings (ΔH_sur) would also be equal in magnitude but opposite in sign to the change in internal energy of the system:
ΔH_sur = -ΔU_sur = -279 kJ
So, the change in internal energy of the surroundings (ΔU_sur) and the change in enthalpy of the surroundings (ΔH_sur) would be 279 kJ and -279 kJ, respectively.
Please note that the sign conventions used for heat and work may vary depending on the specific system of a thermodynamic process, and the above explanation assumes a specific sign convention as mentioned.