Question

One mole of an ideal gas is allowed to extend reversibly and adiabatically from a temperature of $27^\circ$ C. If the work done during the process is $3kJ$ , the final temperature will be equal to: $\left( {{C_v} = 20J{K^ - }} \right)$
A) $150K$
B) $100K$
C) $26.85K$
D) $295K$

Answer 1

By the first law of thermodynamics we know that $\Delta U = q + w$ and for the adiabatic process $q = 0$ . We know $\Delta U = w$ . As we know, $dU = {C_v}dT$ . So the whole equation now becomes $w = {C_v}dT$ . After substituting all values in this equation we get the answer.

Complete answer:
The first law of thermodynamics states that the energy of an isolated system is constant. In other words, this is the law of conservation of energy that is that energy can neither be created nor be destroyed.
$\Delta U = q + w$
Here $\Delta U$ is internal energy change, $q$ heat, and $w$ is work.
In question, it is given that the process is reversible and adiabatic and we know that in an adiabatic process $q = 0$ . It means that there is no heat change between the system and the surrounding.
Now, $\Delta U = w$
We know that at a constant volume heat capacity $\Delta U = {C_v}dT$
Rearranging the equation we get
$w = {C_v}dT$
Here $dT$ is the temperature change. We know that in expansion $\left( {{T_1} > {T_2}} \right)$
$w = {C_v}\left( {{T_1} - {T_2}} \right)$
Now,
Given information is $w = 3kJ = 3000J$
${C_v} = 20J/K$
For changing the temperature from Celsius to kelvin add $273$
${T_1} = 27 + 273$
${T_1} = 27^\circ C = 300K$
${T_2} = ?$
Now upon substitution of all values in the above equation we get:
$3000 = 20\left( {300 - {T_2}} \right)$
After rearranging the equation
${T_2} = 300 - 150$
${T_2} = 150K$
Hence the correct option is (A).

Note:
For changing the temperature from Celsius to kelvin add $273$ . Here in the question work is given in kilojoule and ${C_v}$ value is given in joule per kelvin so don’t forget to make all units same, for this change kilojoule into joule otherwise we will end up with the wrong answer. For an adiabatic change, heat change will be zero.