Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process.
In the light of the above statement, choose the correct answer from the options given below :
A.Both (A)$(A)$ and (R)$(R)$ are true and (R)$(R)$ is the correct explanation of (A)$(A)$
B.(A)$(A)$ is true but (R)$(R)$ is false
C.(A)$(A)$ is false but (R)$(R)$ is true
D.Both (A)$(A)$ and (R)$(R)$ are true but (R)$(R)$ is NOT the correct explanation of (A)$(A)$
Solution & Explanation
### Related Formula
T_1 V_1^gamma-1 = T_2 V_2^gamma-1$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$
### Core Logic
Assertion (A) review:
An insulated container means the process is adiabatic (Q=0$Q=0$). When the gas is shrunk (compressed) to half its volume (V_2 = V_1 / 2$V_2 = V_1 / 2$),
T_2 = T_1 left(fracV_1V_2
ight)^gamma-1 = T_1 (2)^gamma-1$$T_2 = T_1 \left(\frac{V_1}{V_2}
ight)^{\gamma-1} = T_1 (2)^{\gamma-1}$$
Since gamma > 1$\gamma > 1$, T_2 > T_1$T_2 > T_1$. Therefore, temperature increases during adiabatic compression, which makes Assertion (A) false.
Reason (R) review:
Free expansion occurs when a gas expands into a vacuum inside an insulated container. No work is done (W=0$W=0$) and no heat is exchanged (Q=0$Q=0$), hence it is adiabatic. It cannot spontaneously reverse, so it is irreversible. Thus, Reason (R) is true.
### Pattern Recognition
Adiabatic compression always raises temperature due to work being done on the gas, while free expansion keeps the temperature of an ideal gas constant (dI=0$dI=0$ as W=0, Q=0$W=0, Q=0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Keywords:#Adiabatic compression temperature#JEE Main 2025 Evening Q6#Thermodynamics JEE Main 2025#Free expansion ideal gas
More Thermodynamics Previous-Year Questions — Page 6
Q38jee_main_2024_31_jan_morningIsobaric Process
The given figure represents two isobaric processes for the same mass of an ideal gas, then
A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.
### Related Formula
PV = nRT$PV = nRT$
### Core Logic
From the Ideal Gas Law:
V = left(fracnRPright) T$$V = \left(\frac{nR}{P}\right) T$$
In a V-T$V-T$ graph, the equation of the line represents y = mx$y = mx$, where the slope m$m$ is:
textSlope = fracnRP$$\text{Slope} = \frac{nR}{P}$$textSlope propto frac1P$$\text{Slope} \propto \frac{1}{P}$$
Thus, a higher slope corresponds to a lower pressure.
### Step 2: Compare Slopes
From the given figure, the slope of line 2 is greater than the slope of line 1:
(textSlope)_2 > (textSlope)_1$$(\text{Slope})_2 > (\text{Slope})_1$$
Therefore, inversely:
P_2 < P_1$P_2 < P_1$
or P_1 > P_2$P_1 > P_2$.
### Pattern Recognition
In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
More Thermodynamics Questions — jee_main_2025_24_jan_evening
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