Question

A rigid container contain 1kg ${N_2}$ gas at 400K if final temperature is 800K then find change in entropy $\left( {{\mathbf{J}}{{\mathbf{K}}^{ - 1}}} \right)$
[Given:${C_p} = 29.099\;joule$]
(a) 512.86
(b) 1025.736
(c) 256.43
(d) 170.9

Answer 1

Hint: At constant pressure heat change can be defined as its enthalpy change of the system. We can define heat capacity as the change in temperature in the sample for a given amount of heat.

Complete Step by step answer: As we define the heat capacity, molar heat capacity or specific heat capacity is more specific as they are defined on the mass of the substance. Specific heat capacity can be defined as the energy that must be added to the mole of substance in order to increase the temperature by one unit.

Now let us see what is meant by entropy. Entropy is defined as a measure of randomness or disorder of the system. Change in entropy of a system is given as the ratio of heat absorbed $\left( q \right)$ isothermally and reversibly by the system to the temperature$\left( T \right)$ at which heat is absorbed.
$\vartriangle S = \dfrac{q}{T}$
Where $n$, is the number of moles and $T$is the final temperature.
Now, we can see the molar specific heat at constant pressure
${C_p} = \dfrac{S}{n}$
Where $n$, is the number of moles and $s$ is the entropy
Hence by definition of molar specific heat we can say,
${C_p} = \dfrac{q}{{n\vartriangle T}}$
$q = n{C_p}\vartriangle T$
We have given 1kg of ${N_2}$, Then
$number\;of\;moles = \dfrac{{mass\;of\;{N_2}}}{{molar\;mass\;of\;{N_2}}} = \dfrac{{1000}}{{28}}$
Also, $\vartriangle T = 800 - 400 = 400$
${C_p} = 29.099J$ [Given]
Substituting the equation
$q = \dfrac{{1000}}{{28}} \times 29.099 \times 400$
$ \Rightarrow q = 415700$
Now from the equation of change in entropy,
$\vartriangle S = \dfrac{q}{T}$
$ \Rightarrow \vartriangle S = \dfrac{{415700}}{{800}}$
$\vartriangle {\text{S}} = 519.625\;J{K^{ - 1}}$
Hence, the closest option is 512.86

Hence the, option (a) is correct.

Note: The entropy change during an adiabatic reversible change is always zero. Since we know that the heat for an adiabatic process will be zero. So we can say that the change in entropy will also be zero.
$q = 0\; \Rightarrow \;\vartriangle S = 0$