An ideal gas undergoes adiabatic expansion against external pressure. Which of the following are correct ?

Internal energy of the gas remains unchanged

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Internal energy of the gas decrease

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Temperature of the system decreases

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ E + P Δ V = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $Δ E + P Δ V = 0$
For adiabatic process, q = 0.
Hence $Δ E = W = - P Δ V$
Also, $Δ E = C_{v} Δ T = - P Δ V$
Since $Δ V$ is positive for expansion, $Δ E$ will be negative and so also $Δ T$

Suggest Corrections

Similar questions

In adiabatic irreversible expansion of an ideal gas against a constant external pressure $P_{2},$ the P - V work is given by ?

2

d m^{3}

5

1

p_{0}

3

s^{\frac{2}{3}} P_{0}

2 {d m}^{3}

5 b a r

1 b a r

Join BYJU'S Learning Program