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Friday, 9 October

JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 9 January 2020, Shift 2, Physics Q21

Question 21 of 75 in this shift, Physics question 21 of 25, Section B.

Starting at temperature 300 K, one mole of an ideal diatomic gas (γ=1.4\gamma = 1.4) is first compressed adiabatically from volume V1V_1 to V2=V116V_2 = \frac{V_1}{16}. It is then allowed to expand isobarically to volume 2V22V_2. If all the processes are the quasi-static then the final temperature of the gas (in °K) is (to the nearest integer) ______.

Official answer

1816.00 to 1820.00

NTA final key (Jan 2020). Any value in the range scored.

Chapter
Thermodynamics
Topic
Thermodynamic Processes
Idea tested
Adiabatic T-V and P-T Relations

Same idea in other shifts

Asked 3× in all
  1. 25 Jul 2021, Shift 1 · Q20A monoatomic ideal gas, initially at temperature T1T_1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to…EasySingle correct
  2. 2 Sep 2020, Shift 1 · Q22An engine takes in 5 moles of air at 20∘20^\circC and 1 atm, and compresses it adiabaticaly to 1/10th1/10^{\mathrm{th}} of the original volume.…MediumNumerical value

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 42 of the 174 JEE Main shifts held from 2017 to 2026.