Thermodynamics's Question 54 | Chemistry XI - Part 1 | NCERT Exemplar Questions

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Subjective

Match the following

A.	Entropy of vaporisation	1.	decreases
B.	K for spontaneous process	2.	is always positive
C.	Crystalline solid state	3.	lowest entropy
D.	$$\Delta U$$ in adiabatic expansion of ideal gas	4.	$$\frac{\Delta H_{vap}}{T_b}$$

Explanation

A. $\rightarrow(2,4)$

B. $\rightarrow$ (2)

C. $\rightarrow$ (3)

D. $\rightarrow$ (1)

A. Entropy of vaporisation is always positive. It is equal to $\Delta H_{\text {vap }} / T_b$

B. $\Delta_r G^{\circ}=-R T \ln K$

If $K$ is positive, $\Delta_r G^{\circ}=$ negative and reaction is spontaneous.

C. Crystalline solid state has lowest entropy.

D. During adiabatic expansion of an ideal gas, $q=0$. Hence, $\Delta U=q+W$ gives $\Delta U=W$, i.e., work done is at the cost of internal energy which decreases.

MCQ (Single Correct Answer)

Assertion (A) Combustion of all organic compounds is an exothermic reaction.

Reason (R) The enthalpies of all elements in their standard state are zero.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$

Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$

$A$ is true but $R$ is false

$A$ is false but $R$ is true

MCQ (Single Correct Answer)

Assertion (A) Spontaneous process is an irreversible process and may be reversed by some external agency.

Reason (R) Decrease in enthalpy is a contributory factor for spontaneity.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$

Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$

$A$ is true but $R$ is false

$A$ is false but $R$ is true

MCQ (Single Correct Answer)

Assertion (A) A liquid crystallises into a solid and is accompanied by decrease in entropy. Reason (R) In crystals, molecules organise in an ordered manner.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$

Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$

$A$ is true but $R$ is false

$A$ is false but $R$ is true

Subjective

Derive the relationship between $\Delta H$ and $\Delta U$ for an ideal gas. Explain each term involved in the equation.

Explanation

From the first law of thermodynamics, $\quad q=\Delta U+p \Delta V$

If the process carried out at constant volume, $\Delta V=0$

Hence, $$q_v=\Delta U$$

[Here, $q_v=$ Heat absorbed at constant volume, $\Delta U=$ change in internal energy]

Similarly, $q_p=\Delta H$

Here, $\quad q_p=$ heat absorbed at constant pressure

$\Delta H=$ enthalpy change of the system.

Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure.

As we know that at constant pressure, $\Delta H=\Delta U+p \Delta V$ where, $\Delta V$ is the change in volume.

This equation can be rewritten as $\Delta H=\Delta U+p\left(V_f-V_i\right)=\Delta U+\left(p V_f-p V_i\right)\quad \text{... (i)}$

where, $$V_i$$ = initial volume of the system $$V_f$$ = final volume of the system

But for the ideal gases, $$p V=n R T$$

So that $$p V_1=n_1 R T$$

and $$p V_2=n_2 R T$$

where, $n_1=$ number of moles of the gaseous reactants

$n_2=$ number of moles of the gaseous products.

Substituting these values in Eq. (i), we get

$$\begin{aligned} & \Delta H=\Delta U+\left(n_2 R T-n_1 R T\right) \\ \text{or}\quad & \Delta H=\Delta U+\left(n_2-n_1\right) R T \end{aligned}$$

where, $\Delta n_g=n_2-n_1$ is the difference between the number of moles of the gaseous products and gaseous reactants.

Putting the values of $\Delta H$ and $\Delta U$ we get

$$q_p=q_v+\Delta n_g R T$$

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