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Question: If 0.1kcal of heat is added to 1.2L of \({O_2}\) in a cylinder at constant pressure of 1 atm, the vo...

If 0.1kcal of heat is added to 1.2L of O2{O_2} in a cylinder at constant pressure of 1 atm, the volume increases to 1.5L. Calculate ΔU\Delta U and ΔH\Delta H of the process (1L-atm = 100J).

Explanation

Solution

When heat is added to the system at constant pressure, the change in enthalpy is equal to the heat flow. The internal energy increases when heat is provided to the system.

Formula used: ΔU=q−PΔV\Delta U = q - P\Delta V and ΔH=qp\Delta H = {q_p}

Complete step-by-step solution:
As we know, change in internal energy of a system is the sum of heat and work. The quantity of heat transferred from the surroundings to the system is q and the work done in the process is w, then the change in internal energy is given by
ΔU=Q+W\Delta U = Q + W
In the above equation, work done can be represented as expansion and compression of gases, written as
W=PΔVW = P\Delta V
Looking at the change in volume, we can determine whether it is expansion or compression.
In a closed container, if a piston is being pushed down by supplying energy, compression of the gas occurs resulting in a decrease in volume. This tells us that work is done on the system and it is positive. And if the volume increases on expansion of gas against an external pressure and the piston comes up, that means work is done by the system and it is negative.
Here, as the volume is increasing from 1.2L to 1.5L, we say, work is done by the system.
Therefore, the formula of change in internal energy now becomes
ΔU=q−PΔV\Delta U = q - P\Delta V
Given: heat (q)=1 kcal i.e. 1000cal
Initial volume (V1)=1.2L({V_1}) = 1.2L
Final volume (V2)=1.5L({V_2}) = 1.5L
pressure (P) =1 atm
Putting the given values in the above formula, we get
ΔU=1000cal−[1atm(1.5−1.2)L]\Delta U = 1000cal - [1atm(1.5 - 1.2)L]
⇒\Rightarrow ΔU=992.7cal or 0.993kcal\Delta U = 992.7cal \, or \, 0.993kcal
Now, we have to calculate ΔH\Delta H
At constant pressure,
We know that, ΔH=qp\Delta H = {q_p}
We know that, ΔH=qp\Delta H = {q_p}
Therefore, ΔH=1kcal\Delta H = 1kcal

Hence, for the given system, ΔU=992.7cal or 0.993 kcal\Delta U = 992.7cal \,or \, 0.993\,kcal and ΔH=1kcal\Delta H = 1kcal

Note: The change in internal energy states the first law of thermodynamics. According to this law, energy can neither be created nor be destroyed; it can only be transformed from one form to another. Applications of this law include adiabatic, isochoric, isothermal, isobaric and free expansion processes.