JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 26 February 2021, Shift 2, Physics Q23

Question 23 of 90 in this shift, Physics question 23 of 30, Section B.

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V = K

T^{\frac{2}{3}}

. The workdone when temperature changes by 90 K will be xR. The value of

x

is ________. [R = universal gas constant]

Official answer

NTA final key.

Chapter: Thermodynamics
Topic: Thermodynamic Processes
Idea tested: Polytropic Process Analysis

Same idea in other shifts

Asked 5× in all

1 Feb 2024, Shift 1 · Q38The pressure and volume of an ideal gas are related as $PV^{\frac{3}{2}} = K$ (Constant). The work done when the gas is taken from state A…MediumSingle correct
30 Jan 2024, Shift 1 · Q38Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are $C_A$ and $C_B$ . The molar heat…HardDiagram basedHas a figure
25 Feb 2021, Shift 1 · Q29In a certain thermodynamical process, the pressure of a gas depends on its volume as $kV^{3}$ . The work done when the temperature changes…MediumNumerical value
25 Feb 2021, Shift 2 · Q5Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If $V_2 = 2V_1$ then the ratio of temperature…EasyDiagram basedHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.