
Recall the formula of the first law of thermodynamics in terms of change in internal energy and work done by the system, ΔU=Q−W, given, ΔU=2500J, Q=606000×t=(100t)J and W=P×t=90t,
⇒2500J=100tJ−90tJ⇒t=2.5×102s.
JEE Main 2021 — Physics Thermodynamics
An electric appliance supplies 6000Jmin−1, heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5×103J?
Held on 26 Aug 2021 · Verified 6 Jul 2026.
2.5×101s
2.5×102s
2.4×103s
4.1×101s
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
10 mole of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The change in the internal energy of the gas is $\_\_\_\_$ cal. (The molecular specific heat of oxygen at constant pressure, $C_{P}=7 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C}$ and $\mathrm{R}=2 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C}$.)
An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 = $ constant. The coefficient of volume expansion of the gas is _______.
In the following $p-V$ diagram the equation of state along the curved path is given by $(V-2)^{2}=4 a p$ where $a$ is a constant. The total work done in the closed path is 
The volume of an ideal gas increases 8 times and temperature becomes $(1 / 4)^{\text {th }}$ of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta \mathrm{Q}=0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
A certain gas is isothermally compressed to $\left(\dfrac{1}{3}\right)^{rd}$ of its initial volume ($V_o = 3$ litre) by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5$ N/m$^2$, the magnitude of work done on the gas is _______ J.
Work through every JEE Main Thermodynamics PYQ, year by year.