Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

An ideal gas undergoes a cyclic process as shown in Figure.

$\Delta U_{BC}= -5 kJ mol^{-1}, q_{AB} = 2kJ mol^{-1}$

$W_{AB}= -5 kJ mol^{-1}, W_{CA} = 3kJ mol^{-1}$

Heat absorbed by the system during process CA is :

Option 1)
$-5kJ mol^{-1}$

Option 2)
$+5kJ mol^{-1}$

Option 3)
$18kJ mol^{-1}$

Option 4)
$-18kJ mol^{-1}$

Answers (1)

As we learned

Isochoric Process -

Volume is constant during the process

$dV= 0$

$\Delta V= 0$

- wherein

Isobaric process -

Pressure is constant during the process

$dP= 0$

$\Delta P= 0$

- wherein

$AB \rightarrow$ Isobaric

$BC \rightarrow$ Isochoric

$CA \rightarrow$ Not Defined

$\Delta U_{nB}= q+w$

2 - 5 = -3KJ

$\Delta U_{ABC} =\Delta U_{AB}+\Delta U_{BC}$

-3 - 5 = -3KJ

$\Delta U_{CBA} =+8$

Q+w

8 = Q+3

Q= +5KJ

Option 1)

$-5kJ mol^{-1}$

Option 2)

$+5kJ mol^{-1}$

Option 3)

$18kJ mol^{-1}$

Option 4)

$-18kJ mol^{-1}$

Posted by

subam

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks

anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?

anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question