CBSE Class 11

Conducting Body: Individual CBSE affiliated schools
Who It Is For: Students at the end of the Class 11 academic year
Purpose: Promotion to Class 12
Subjects: Varies by stream (Science, Commerce, Humanities)
Question Types: Objective, short answer, and long answer questions
Duration: Typically 3 hours per paper
Frequency: Annually, at the end of the academic session

The CBSE Class 11 final exam is a school-level examination conducted at the end of the academic year. It is taken by students who have completed Class 10 and have entered their first year of senior secondary education, having chosen a specific stream like Science, Commerce, or Humanities. The primary purpose of this exam is to assess a student's understanding of the year's curriculum and determine their promotion to Class 12. The results serve as an important benchmark for a student's preparation for the Class 12 board exams.

The papers are prepared and evaluated internally by the student's own school, although they must adhere to the syllabus and question paper design prescribed by the CBSE. The main difficulty students face is the significant increase in syllabus depth and complexity compared to Class 10. Each stream introduces advanced topics and requires a more analytical approach to learning. Managing the volume of information across five or six subjects demands consistent effort and effective time management throughout the year, making it a challenging transition for many.

How a question gets worked here

From a physics paper rather than CBSE Class 11, worked the way every question on this site is. CBSE Class 11 questions are being added.

JEE Advanced 2026

Ten moles of an ideal monoatomic gas, initially in state $a$ at atmospheric pressure and temperature $T_a=27^\circ\text{C}$ , is enclosed in a metal cylinder of volume $V_0$ fitted with a frictionless piston. The gas is suddenly compressed to state $b$ with volume $V_0/3$ . Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature $11^\circ\text{C}$ until the gas reaches the temperature of the water bath, which is denoted as state $c$ . Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state $f$ . If $R$ is universal gas constant, then the correct option(s) is/are: [Given: $9^{1/3}=2.08$ ]

Thermodynamics Ten moles of an ideal monoatomic gas, initially in state

a

at atmospheric pressure and temperature

T_a=27^\circ\text{C}

, is enclosed in a metal cylinder of volume

V_0

fitted with a frictionless piston. The gas is suddenly compressed to state

b

V_0/3

. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature

11^\circ\text{C}

until the gas reaches the temperature of the water bath, which is denoted as state

c

. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state

f

R

is universal gas constant, then the correct option(s) is/are: [Given:

9^{1/3}=2.08

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Identify the parameters of the gas and the states.

For a monoatomic ideal gas, the heat capacity at constant volume $C_V$ and the adiabatic index $\gamma$ are:

C_V = \frac{3}{2}R

\gamma = \frac{5}{3}

The initial state $a$ parameters are:

n = 10\text{ mol}

T_a = 27^\circ\text{C} = 300\text{ K}

V_a = V_0

Step 2: Analyze the adiabatic compression from state $a$ to state $b$ .

Since the gas is suddenly compressed, the process $a \to b$ is adiabatic. The volume at state $b$ is $V_b = \frac{V_0}{3}$ .

Find the temperature $T_b$ using the adiabatic relation:

T_b = T_a \left(\frac{V_a}{V_b}\right)^{\gamma - 1}

T_b = 300 \times 3^{2/3}

Using the given value $9^{1/3} = 3^{2/3} = 2.08$ :

T_b = 300 \times 2.08 = 624\text{ K}

Step 3: Calculate the change in internal energy for the process $a \to b$ .

\Delta U_{a\to b} = n C_V (T_b - T_a)

\Delta U_{a\to b} = 10 \times \frac{3}{2}R \times (624 - 300)

\Delta U_{a\to b} = 15R \times 324 = 4860R

Thus, option (2) is correct.

Step 4: Analyze the processes $b \to c$ and $c \to f$ .

In process $b \to c$ , the piston is stationary (isochoric process) and the gas cools to the water bath temperature:

T_c = 11^\circ\text{C} = 284\text{ K}

In process $c \to f$ , the gas undergoes isothermal expansion back to the initial volume:

T_f = T_c = 284\text{ K}

V_f = V_0

Step 5: Calculate the net change in internal energy for the entire process.

Since internal energy is a state function, the net change depends only on the initial and final temperatures:

\Delta U_{\text{net}} = n C_V (T_f - T_a)

\Delta U_{\text{net}} = 10 \times \frac{3}{2}R \times (284 - 300)

\Delta U_{\text{net}} = 15R \times (-16) = -240R

Thus, option (3) is correct.

Step 6: Calculate the pressure at state $b$ .

Using the adiabatic relation for pressure:

P_b = P_a \left(\frac{V_a}{V_b}\right)^{\gamma}

P_b = 1\text{ atm} \times 3^{5/3}

P_b = 3 \times 3^{2/3} = 3 \times 2.08 = 6.24\text{ atm}

Since $P_b = 6.24\text{ atm} \neq 2.08\text{ atm}$ , option (4) is incorrect.

Step 7: Verify the $P-V$ diagram.

The processes are:

$a \to b$ : Adiabatic compression (curve rising to the left)
$b \to c$ : Isochoric cooling (vertical line downwards)
$c \to f$ : Isothermal expansion (curve going to the right)

This sequence matches the described $P-V$ diagram, so option (1) is correct.

Answer: The correct options are (1), (2), and (3).

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