Q.

Calculate the work done by the system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas (γ=4/3) from 300 K and pressure 10 atm to 1 atm : (in KJ)

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

detailed_solution_thumbnail

In irreversible adiabatic process work done

w=-pext v2-v1 (or) w=nRγ-1T2-T1

For irreversible processpvr is not valid and final temperature is to be obtained by equating the above two expressions.

  So, -p2v2-v1=nRγ-1T2-T1 P2nT1RP1-nT2RP2=nRγ-1T2-T1 P2T1P-T2=1γ-1T2-T1  so, T1P2P1+1γ-1,T21+1γ-1 T2=T1γ-1γP2P1+1γ-1  =3001/34/3110+11/3  =300×14×3110=232.5 w=2×8.314×10-31/3[232.5-300]; so, ans 3

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon