
W=21πR2=21×π×(2200×103)×2200×10−6=210π=5π J
JEE Main 2025 — Physics Thermodynamics
The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) : 
Held on 24 Jan 2025 · Verified 6 Jul 2026.
5π
40π
10π
zero
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10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to $P_{2}$ is $\alpha$ Joule ($P_{1}=21.7 \mathrm{~Pa}$ and $\left.P_{2}=30 \mathrm{~Pa}, \mathrm{C}_{v}=21 \mathrm{~J} / \mathrm{K}. \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol}. \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$. 
Heat is supplied to a diatomic gas at constant pressure. Then the ratio of $\Delta Q : \Delta U : \Delta W$ is _______.
The r.m.s. speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$. (Mass of oxygen molecule/mass of hydrogen molecule $=32 / 2$)
Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^{\circ} \mathrm{C}$ and $40^{\circ} \mathrm{C}$ respectively. Given the thermal conductivity of $\operatorname{rod} x$ is three times of that of $\operatorname{rod} y$, the temperature at junction points $B$ and $E$ are (close to): 
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Statement I: Change in internal energy of a system containing $n$ mole of ideal gas can be written as $\Delta U = n C_v (T_f - T_i) = \dfrac{nR}{\gamma - 1}(T_f - T_i)$, where $\gamma = \dfrac{C_p}{C_v}$, $T_i =$ initial temperature, $T_f =$ final temperature. Statement II: Relation between degree of freedom $f$ and $\gamma (= C_p/C_v)$ is $\left(\gamma = 1 + \dfrac{2}{f}\right)$ Choose the correct answer from the options given below
Work through every JEE Main Thermodynamics PYQ, year by year.