Solveeit Logo
Login

Question

Question: If \(1.0\) kcal of heat is added to \(1.2\) L of \({{\text{O}}_{\text{2}}}\) in a cylinder of consta...

If 1.01.0 kcal of heat is added to 1.21.2 L of O2{{\text{O}}_{\text{2}}} in a cylinder of constant pressure of 1 atm, the volume increases to 1.51.5 L. Calculate ΔE\Delta {\text{E}} and ΔH\Delta {\text{H}} of the process:
A.ΔE = 0.993kcal\Delta {\text{E = 0}}{\text{.993kcal}} , ΔH = 1kcal\Delta {\text{H = 1kcal}}
B.ΔE = - 0.993kcal\Delta {\text{E = - 0}}{\text{.993kcal}} , ΔH = 1kcal\Delta {\text{H = 1kcal}}
C.ΔE = 0.993kcal\Delta {\text{E = 0}}{\text{.993kcal}} , ΔH = - 1kcal\Delta {\text{H = - 1kcal}}
D.None of these

Explanation

Solution

From the first law of thermodynamics, we know that the change in internal energy is equal to the sum of heat added to the system and the work done on the system.
The change in energy at constant pressure, i.e., enthalpy change is equal to the sum of the change in internal energy at constant volume, i.e., internal energy change and the pressure volume energy of the system.

Complete step by step answer:
The mathematical form of the first law of thermodynamics is given by the following expression:
ΔE = q + w\Delta {\text{E = q + w}}
Here, ΔE\Delta {\text{E}} is the change in internal energy of the system, q is the heat supplied to the system and w is the work done on the system.
If the work done by the system is only pressure – volume work, then, the work done is given by:
w=−PΔV{\text{w}} = - {\text{P}}\Delta {\text{V}}
Here, P is the constant pressure and ΔV\Delta {\text{V}} is the change in volume.
Then, the change in internal energy will be:
ΔE = q−PΔV\Delta {\text{E = q}} - {\text{P}}\Delta {\text{V}}
According to the given question, q=1.0kcal{\text{q}} = 1.0{\text{kcal}} , P is 1 atm.
ΔV = 1.5 - 1.2 ⇒ΔV = 0.3L  \Delta {\text{V = 1}}{\text{.5 - 1}}{\text{.2}} \\\ \Rightarrow \Delta {\text{V = 0}}{\text{.3L}} \\\
The work done is:
w=−1atm×0.3L ⇒w=−0.3Latm  {\text{w}} = - 1{\text{atm}} \times 0.3{\text{L}} \\\ \Rightarrow {\text{w}} = - 0.3{\text{Latm}} \\\
Now, 1Latm = 101.3J1{\text{Latm = }}101.3{\text{J}}.
Thus,
w=−0.3×101.3J ⇒w=−30.39J {\text{w}} = - 0.3 \times 101.3{\text{J}} \\\ \Rightarrow {\text{w}} = - 30.39{\text{J}} \\\
Again, 4.18J=1cal4.18{\text{J}} = 1{\text{cal}} .
So,
w=−30.394.18cal ⇒w=−7.27cal ⇒w=−0.007kcal  {\text{w}} = \dfrac{{ - 30.39}}{{4.18}}{\text{cal}} \\\ \Rightarrow {\text{w}} = - 7.27{\text{cal}} \\\ \Rightarrow {\text{w}} = - 0.007{\text{kcal}} \\\
Finally,
ΔE = 1.0kcal - 0.007kcal ⇒ΔE = 0.993kcal  \Delta {\text{E = 1}}{\text{.0kcal - }}0.007{\text{kcal}} \\\ \Rightarrow \Delta {\text{E = }}0.993{\text{kcal}} \\\
Now, the enthalpy change is given by:
ΔH = ΔE+PΔV\Delta {\text{H = }}\Delta {\text{E}} + {\text{P}}\Delta {\text{V}}
Substitute the values obtained above in this equation and we will have:
ΔH = 0.993 + 0.007 ⇒ΔH = 1kcal  \Delta {\text{H = 0}}{\text{.993 + 0}}{\text{.007}} \\\ \Rightarrow \Delta {\text{H = }}1{\text{kcal}} \\\

So, the correct option is A.

Note: If a process takes place at constant volume, then the change in volume ΔV = 0\Delta {\text{V = 0}}. Then, PΔV = 0{\text{P}}\Delta {\text{V = 0}} and so ΔE = q\Delta {\text{E = q}} . This indicates that the internal energy change at constant volume and constant temperature is equal to the heat evolved or absorbed.
The enthalpy change can also be found out by an alternate relation:
ΔH = ΔE+ΔngRT\Delta {\text{H = }}\Delta {\text{E}} + \Delta {{\text{n}}_{\text{g}}}{\text{RT}}
Here, Δng\Delta {{\text{n}}_{\text{g}}} is the change in the number of moles of gaseous products and reactants, R and T is the gas constant and temperature respectively.