Question

A gas is confined to a cylinder under constant atmospheric pressure,. When the gas undergoes a particular chemical reaction , it releases 135 kJ of heat to its surroundings and does 63 KJ of P-V Work on its surroundings. What are the values of ∆H and ∆E for this process ?

Answer 1

`\begin{gathered} \triangle H=-135kJ \\ \triangle E=-198kJ \end{gathered}`

Explanations:

From the question, we are given the following

Amount of heat released to the surroundings = 135kJ

Work done to its surroundings q = 63 kJ

The derivation of the enthalpy at constant pressure is expressed as;

`\triangle H=\triangle U+\triangle(P_(int)V)`

where;

`\begin{gathered} \triangle U\text{ is the internal energy} \\ P_{int_{}}\text{ is the internal pressure} \\ V\text{ is the volume of the gas} \end{gathered}`

Since the cylinder gas is under constant pressure, then the enthalpy will be equal to the work done to have:

`\triangle H=q`

Since q = 63kJ, hence;

`\triangle H=q=-135kJ`

Next is to calculate the change in the change in the internal energy

Using the law of energy conservation which states that the change in internal energy is equal to the heat transferred to, less the work done by, the system. Mathematically;

`\begin{gathered} \triangle E=q+W \\ \triangle E=-135kJ+(-63kJ) \\ \triangle E=-198kJ \end{gathered}`

Hence the change in internal energy for the process is -198kJ