A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy \Delta U of the gas in joules will be

  • Option 1)

    -500 J

  • Option 2)

    -505 J

  • Option 3)

    +505 J

  • Option 4)

    1136.25 J

 

Answers (1)
V Vakul

As we learnt

Work Done for Irreversible Isothermal Expansion of an ideal gas -

W= -P_{ext}\left ( V_{f} -V_{i}\right )

- wherein

P_{ext} may be equal to P_{f} or may not be equal to P_{f}, but work done is always calculated by P_{ext}

 

 and

Adiabatic Process -

Heat exchange between system and surrounding is zero i.e. q= 0
 

- wherein

\Delta E= q+w

q= 0

\Delta E=w

 

 Work done in this process will be 

W = -P_{ext}^{\Delta V}

= -2.5[4.5 - 2.5] = -5 L atm

= - 5 101.3 J = -206J

since system is well insuvated

q = 0

=>\Delta U_W = 0

or \Delta U = +W = -505 J

 


Option 1)

-500 J

this is the incorrect option

Option 2)

-505 J

this is the correct option

Option 3)

+505 J

this is the incorrect option

Option 4)

1136.25 J

this is the incorrect option

Preparation Products

Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout NEET May 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 4999/-
Buy Now
Knockout NEET May 2022 (Subscription)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 5499/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions