One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27 °C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27 °C doing the same amount of work W, then its fin

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27 $° C$ and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27 $° C$ doing the same amount of work W, then its final temperature will be (close to) ______ $° C$ .

$(\log ₑ 2 = 0.693)$ [2026]

1 −189

2 −117

3 −30

4 −56

Ans.
(4)

$For isothermal process$

$W_{isothermal} = n R T l n (\frac{V_{2}}{V_{1}})$

$= 1 \cdot R \cdot 300 \cdot l n (2)$

$= 300 R (0.693) \dots (1)$

$Now for adiabatic process,$

$It is given work done in isothermal = work done in adiabatic$

$W_{adiabatic} = \frac{n R (T_{1} - T_{2})}{γ - 1} \dots (2)$

$(1) = (2)$

$\frac{n R (300 - T_{final})}{1.4 - 1} = 300 R (0.693)$

$T_{final} = 216.84 K$

$= - 56.3 ° C$

Want Us to Email you About Special Offers & Updates?