Question

For reversible and irreversible adiabatic
expansion of ideal gas incorrect statement
is
1 (Tp)rev < (Tp) ir
2 (Pf)rev < (Pt)irr
3 (Vf)rev < (Vf) irr
4 (Vprev> (VF) ir
​

Answer 1

Answer:

3 (Vf)rev < (Vf) irr

Explanation:

humm...........

Answer 2

Answer:

According to first law of thermodynamics

ΔQ=ΔU+ΔW

An isolated system is adiabatic. This means ΔQ=0. The first law in this case yields

0=ΔU+ΔW=>ΔW=−ΔU … … (1)

For expansion, ΔW is positive and hence ΔU is negative. This means T

f

is less than T

i

in both the cases. However it is interesting to see in which case the final temperature is greater.

For the same expansion of volume, the work done in irreversible process is greater than that in reversible one because the system has to work against friction etc. Thus

ΔW

irreversible

>ΔW

r

eversible

=>−ΔU

irreversible

>−ΔU

reversible

(from equation (1))

=>ΔU

irreversible

<ΔU

reversible

=>ΔT

irreversible

<ΔT

reversible

=>T

f−irreversible

>T

f−reversible