Question

An ideal gas undergoes isothermal compression from ${\mathbf{5}}{{\mathbf{m}}^{\mathbf{3}}}$ to ${\mathbf{1}}{{\mathbf{m}}^{\mathbf{3}}}$ against a constant external pressure of ${\mathbf{4N}}{{\mathbf{m}}^{ - 2}}$ . Heat released in this process is used to increase the temperature of 1 mole of ${\mathbf{Al}}$. If molar heat capacity of ${\mathbf{Al}}\; = \;{\mathbf{24Jmo}}{{\mathbf{l}}^{ - {\mathbf{1}}}}{{\mathbf{k}}^{ - {\mathbf{1}}}}$, the temperature of ${\mathbf{Al}}$ increased by:
A.$\dfrac{3}{2}{\text{K}}$
B.$\dfrac{2}{3}{\text{K}}$
C.$1{\text{K}}$
D.$2{\text{K}}$

Answer 1

To answer this question, you should recall the laws of thermodynamics, especially the first law of thermodynamics. You should be able to calculate work done by gas.
Formula used:

1. ${\text{W}} = - {{\text{P}}_{{\text{ext}}}}({{\text{V}}_{\text{2}}} - {{\text{V}}_{\text{1}}})$
   where W =Work Done on isothermal irreversible for an ideal gas,
   ${{\text{P}}_{{\text{ext}}}}$ = External Pressure and ${{\text{V}}_{\text{2}}} - {{\text{V}}_{\text{1}}}$ is the change in volume
2. ${{\Delta U }} = {{ }}\Delta {\text{Q }} - {{ }}\Delta {\text{W}}$
   where ${{\Delta U}}$ is the change in internal energy of the system, ${{\Delta Q }}$is heat transfer and ${{\Delta W}}$is work done on the system
3. ${\text{q = n}}{{\text{C}}_{\text{m}}}{{\Delta T}}$
   where q is the heat applied, ${{\text{C}}_{\text{m}}}$ is the molar heat capacity, n is the mole and $\Delta {\text{T}}$ is the change in temperature.

Complete step by step answer:
First law of thermodynamics, is more popularly known as the law of conservation of energy states that energy can neither be created nor destroyed, but it can be changed from one form to another. The limitation of this first law of thermodynamics is that it fails to explain why heat flows from hot end to cold end when a metallic rod is heated at one end not in other cases and vice-versa. This means that the first law only quantifies the energy transfer that takes place during this process. It is the second law of thermodynamics which provides the criterion for the feasibility of the various processes. Using the values given in the question and substituting in the formula for work done:
${\text{W}} = - {{\text{P}}_{{\text{ext}}}}({{\text{V}}_{\text{2}}} - {{\text{V}}_{\text{1}}})$
Substituting the values, we get:
${\text{W}} = - 4{\text{N}}{\text{.}}{{\text{m}}^{ - 2}} \times \left( {1 - 5} \right){{\text{m}}^3}$
$\Rightarrow {\text{W}} = - 4{\text{N}}{\text{.}}{{\text{m}}^{ - 2}} \times - 4{{\text{m}}^3} = 16N.m$
For the Isothermal process, we know, $\Delta {\text{U}} = 0$.
Hence using the first law of thermodynamics we can say that ${\text{q}} = - 16{\text{J}}$.
This heat will be used to increase temperature.
Using the equation:
${\text{q = n}}{{\text{C}}_{\text{m}}}{{\Delta T}}$
Here, $n = 1$and ${C_{m}} = 24{\text{J/}}\left( {{\text{mol}}{\text{.K}}} \right)$ given in the question.
Substituting and solving we get,
$\Delta {\text{T}} = \dfrac{2}{3}{\text{K}}$.

Hence, the correct option is B.

Note:
We can confuse between different types of reactions. Make sure to remember the difference between isobaric, isochoric, isothermal and adiabatic processes. An isobaric process is one where the pressure of the system (often a gas) stays constant. An isochoric process is defined as the thermodynamic process where the volume of the system stays constant. An adiabatic process is a thermodynamic process in which no heat is exchanged between the system and the surrounding.