Physics Q98 – AP-EAMCET 2026

Home Physics ThermodynamicsQuestion #502

Equal amounts of a diatomic ideal gas are contained in two separate cylinders, P and Q. Cylinder P has a movable piston, while cylinder Q has a fixed piston. Both gases are initially at

273 K

, and the same amount of heat is supplied to each. If the gas in cylinder P shows a temperature rise of

20 K

, then the increase in temperature of the gas in cylinder Q is

Clue 1 of 3: Thermodynamic Processes

Identify the type of thermodynamic process occurring in each cylinder (isobaric for P, isochoric for Q) and recall the relevant formulas for heat supplied and work done.

🧠 Full Solution Path

Step 1: Identify the thermodynamic processes and relevant specific heats✦ Active

Cylinder P has a movable piston, which means the pressure remains constant during the heating process. This is an isobaric process. Cylinder Q has a fixed piston, meaning its volume remains constant. This is an isochoric process.

For a diatomic ideal gas, the molar specific heat at constant volume is $C_{v} = \frac{5}{2} R$ .

The molar specific heat at constant pressure is $C_{p} = C_{v} + R = \frac{5}{2} R + R = \frac{7}{2} R$ .

💡 Teacher's Secret Hint

Remember that for an ideal gas, $C_{p} - C_{v} = R$ (Mayer's relation).

Step 2: Apply the First Law of Thermodynamics to Cylinder P (isobaric process)○ Expand

For an isobaric process, the heat supplied ( $Δ Q_{P}$ ) is given by:

Δ Q_{P} = n C_{p} Δ T_{P}

Given that the temperature rise in cylinder P is $Δ T_{P} = 20 K$ . Substituting the value of $C_{p}$ :

Δ Q_{P} = n (\frac{7}{2} R) (20 K)

Step 3: Apply the First Law of Thermodynamics to Cylinder Q (isochoric process)○ Expand

For an isochoric process, the volume is constant, so the work done by the gas ( $Δ W_{Q}$ ) is zero ( $Δ W_{Q} = P Δ V = 0$ ). According to the First Law of Thermodynamics, $Δ Q_{Q} = Δ U_{Q} + Δ W_{Q}$ .

Therefore, the heat supplied ( $Δ Q_{Q}$ ) is equal to the change in internal energy ( $Δ U_{Q}$ ):

Δ Q_{Q} = Δ U_{Q} = n C_{v} Δ T_{Q}

Substituting the value of $C_{v}$ :

Δ Q_{Q} = n (\frac{5}{2} R) Δ T_{Q}

💡 Teacher's Secret Hint

In an isochoric process, all the heat supplied goes into increasing the internal energy of the gas.

Step 4: Equate the heat supplied and solve for

Δ T_{Q}

○ Expand

The problem states that the same amount of heat is supplied to both cylinders, so $Δ Q_{P} = Δ Q_{Q}$ .

n (\frac{7}{2} R) (20 K) = n (\frac{5}{2} R) Δ T_{Q}

We can cancel $n$ and $R$ from both sides of the equation:

\frac{7}{2} \times 20 = \frac{5}{2} \times Δ T_{Q}

Multiply both sides by 2:

7 \times 20 = 5 \times Δ T_{Q}

140 = 5 \times Δ T_{Q}

Now, solve for $Δ T_{Q}$ :

Δ T_{Q} = \frac{140}{5} = 28 K

💡 Teacher's Secret Hint

Always double-check your arithmetic, especially when dealing with fractions.

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Physics Question 98 – AP-EAMCET 2026

Clue 1 of 3: Thermodynamic Processes

🧠 Full Solution Path