Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 6.
Q1498 (2024)
Standard enthalpy of vapourisation for $\mathrm{CCl_4}$ is $30.5\mathrm{kJ mol^{-1}}$. Heat required for vapourisation of $284\mathrm{g}$ of $\mathrm{CCl_4}$ at constant temperature is ________ kJ. (Given molar mass in g mol-; C = 12, Cl = 35.5)
### Related Formula
$$Q = n \times \Delta H_{\text{vap}}^0 \quad \text{where } n = \frac{\text{Mass}}{\text{Molar Mass}}$$
### Core Logic
First, calculate the molar mass of carbon tetrachloride ($\mathrm{CCl}_4$):
$$\text{Molar mass} = 12 + 4(35.5) = 12 + 142 = 154\text{ g/mol}$$
Next, calculate the total number of moles present in $284\text{ g}$ of the substance:
$$n = \frac{284}{154} \approx 1.844\text{ moles}$$
### Step 1: Enthalpy Calculation
Calculate the total energy required for vaporization:
$$\Delta H = 1.844\text{ mol} \times 30.5\text{ kJ/mol} \approx 56.24\text{ kJ}$$
Rounding to the nearest integer value gives **56**.
### Pattern Recognition
Enthalpy of vaporization is an intensive property given per mole. Scale it linearly by multiplying by the total number of moles to find the total extensive heat required.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q84 (2024)
If three moles of an ideal gas at $300\text{ K}$ expand isothermally from $30\text{ dm}^3$ to $45\text{ dm}^3$ against a constant opposing pressure of $80\text{ kPa}$, then the amount of heat transferred is $\text{\quad\quad}$ J.
### Related Formula
First law of thermodynamics framework:
$$\Delta U = Q + W$$
For an isothermal processes involving ideal gases, internal energy change is zero:
$$\Delta U = 0 \implies Q = -W$$
Irreversible work formula expanding against constant external pressure:
$$W = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)$$
### Step 1: Calculate structural work values
Given values:
$$P_{\text{ext}} = 80\text{ kPa} = 80 \times 10^3\text{ Pa}$$
$$V_1 = 30\text{ dm}^3 = 30 \times 10^{-3}\text{ m}^3$$
$$V_2 = 45\text{ dm}^3 = 45 \times 10^{-3}\text{ m}^3$$
$$\Delta V = (45 - 30) \times 10^{-3} = 15 \times 10^{-3}\text{ m}^3$$
$$W = -80 \times 10^3 \times (15 \times 10^{-3}) = -1200\text{ J}$$
### Step 2: Solve for heat magnitude
$$$Q = -W = -(-1200\text{ J}) = 1200\text{ J}$$$
### Pattern Recognition
Constant opposing pressure indicates an irreversible process path. Use $W = -P\Delta V$ instead of logarithmic integrals.
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q1668 (2024)
Which of the following is not correct?
- $\Delta G \text{ is negative for a spontaneous reaction}$
- $\Delta G \text{ is positive for a spontaneous reaction}$
- $\Delta G \text{ is zero for a reversible reaction}$
- $\Delta G \text{ is positive for a non-spontaneous reaction}$
### Core Logic
According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy ($\Delta G$) dictates the spontaneity of a process.
- If $\Delta G \lt 0$ (negative), the process is spontaneous.
- If $\Delta G \gt 0$ (positive), the process is non-spontaneous.
- If $\Delta G = 0$, the system is in equilibrium (reversible process).
Therefore, the statement "$\Delta G$ is positive for a spontaneous reaction" is factually incorrect.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q83 (2024)
Two reactions are given below:
$$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \Delta \mathrm{H}^{\circ} = -822 \mathrm{kJ/mol}$$
$$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \Delta \mathrm{H}^{\circ} = -110 \mathrm{kJ/mol}$$
Then enthalpy change for following reaction
$$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}$$
### Related Formula
According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.
### Core Logic
Let the given reactions be:
(1) $2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \quad \Delta H_1 = -822 \, \mathrm{kJ/mol}$
(2) $\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \quad \Delta H_2 = -110 \, \mathrm{kJ/mol}$
Target Reaction (3):
$$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}, \quad \Delta H_3 = ?$$
To construct the target reaction:
- We need $3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
- We need $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and $2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).
### Step 1: Calculate Net Enthalpy
Target Reaction (3) = $3 \times (2) - (1)$
$$\Delta H_3 = 3 \times \Delta H_2 - \Delta H_1$$
$$\Delta H_3 = 3(-110) - (-822)$$
$$\Delta H_3 = -330 + 822 = 492 \, \mathrm{kJ/mol}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q1853 (2024)
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path $A\rightarrow B\rightarrow C\rightarrow A$ as shown in the diagram. The total work done in the process is ________ J.
{{IMG1}}
### Related Formula
$$W_{\text{cyclic}} = \text{Area enclosed in P-V graph}$$
### Core Logic
The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph.
Note that the provided graph is Volume ($V$) on the y-axis versus Pressure ($P$) on the x-axis.
The path $A \rightarrow B \rightarrow C \rightarrow A$ is traced in a clockwise direction on the $V$-$P$ graph.
Clockwise on a $V$-$P$ graph corresponds to anti-clockwise on a standard $P$-$V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).
### Step 1: Calculating Area
The enclosed region is a right-angled triangle.
Base of triangle on P-axis = $30 - 10 = 20 \text{ kPa}$
Height of triangle on V-axis = $30 - 10 = 20 \text{ dm}^3$
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
$$\text{Area} = \frac{1}{2} \times 20 \times 20 = 200 \text{ kPa}\cdot\text{dm}^3$$
### Step 2: Unit conversion
$1 \text{ kPa} = 10^3 \text{ Pa}$
$1 \text{ dm}^3 = 1 \text{ Litre} = 10^{-3} \text{ m}^3$
$$W = 200 \times 10^3 \text{ Pa} \times 10^{-3} \text{ m}^3$$
$$W = 200 \text{ J}$$
### Pattern Recognition
$1 \text{ kPa} \cdot 1 \text{ L} = 1 \text{ Joule}$. This direct conversion saves time without converting explicitly to standard SI units ($Pa$ and $m^3$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
