Given that P=kT TP= constant ∴ Volume is constant or isochoric process. $\begin{aligned}
& \therefore \mathrm{W}_{\mathrm{D}}=0 \
& \therefore \mathrm{Q}=\Delta \mathrm{U}
\end{aligned}$ Also temperature increases hence internal energy increases.
JEE Main 2025 — Physics Thermodynamics
An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below:
Held on 24 Jan 2025 · Verified 6 Jul 2026.
E Only
A, B, C, D Only
A, D, E Only
A, C Only
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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 $\_\_\_\_$. 
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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.