JEE Main 30 June 2022 Shift 1 Solved Paper

For monoatomic gas degree of freedom f=3 and γ=‌

5

3

Here for gas,
Initial pressure (P1)=75‌kPa
Initial volume (V1)=1200cm3
Final volume (V2)=150cm3
Final pressure (P2)= ?
For adiabatic process,
P1V1γ=P2V2γ
⇒75×(1200)γ=P2(150)γ
⇒P2=75×(‌

1200

150

)‌

5

3

=75×(8)‌

5

3

=75×32‌kPa
=2400‌kPa
Work done in adiabatic process,
W=‌

P2V2−P1V1

1−γ

=‌

2400×103×150×10−6−75×103×1200×10−6

1−‌ 5 3

=‌

(2400×150−75×1200)×10−3

−‌ 2 3

=‌

−810000×10−3

2

=−405‌kJ