Question

How would you calculate delta E of a gas for a process in which the gas absorbs $35\ J$ of heat and does $8\ J$ of work by expanding.

Answer 1

Hint: By the help of the first law of thermodynamics, we can make a relationship between delta E, heat and work.
$\vartriangle U=q\ +\ w$ both heat and work are path function but here the sum of heat and work are state function i.e., it depends upon only the initial and final state.

Complete step by step answer:
In thermodynamics, several concepts are given by the help of which we can find the spontaneity of the reaction as well as heat change during the reaction. Apart from that we can also find the thermodynamic stability of a product or reactant of a given reaction.
In an exothermic process, gets released by the product molecule during its formation and it gives an idea that product are thermodynamically more stable as compared to the stability of the reactant molecules.
In an endothermic process, energy gets absorbed by the reactant molecule to break down itself into products and it gives an idea that reactants are thermodynamically more stable as compared to the stability of reactant molecules.
By first law of thermodynamics,
$\Delta E=\ q\ +\ w$
Where $\Delta \operatorname{E}\ =$ change in internal energy
$\operatorname{Q}=$ amount of heat absorbed or released
$W=$ work done by system or on system
In the given question, $\operatorname{q}=+35J,\ W=-8J$
$\Delta E=35J+(-8)\operatorname{J}$
$\Delta E=(35-8)J$
$\Delta E=27J$
Final answer:
The final answer is $27\ J$
From the above answer it is clear that the internal energy of the system increases during the reaction as the value of $\Delta E$ is positive.

Note:
By the help of first law, we can explain about the internal energy of the reaction, whether during the reaction internal energy increases or decreases. Similarly the amount of work done can also be calculated by the help of a given formula.