For an isothermal free expansion of an ideal gas into vacuum, which one of the following set correct?

Question

For an isothermal free expansion of an ideal gas into vacuum, which one of the following set correct? (A) Δ U=0, q > 0, w < 0 (B) Δ U =0, q > 0, w =0 (C) Δ U=0, q=0, w=0 (D) Δ U < 0, q=0, w < 0

Tardigrade Admin · Accepted Answer

For isothermal process, Δ T =0 . For an ideal gas, internal energy U depend only on temperature. So, Δ U=0 Now Δ U=q + w q=-w=P Δ U For free expansion P text ext =0 So, q=-w=0 ∴ Δ U=0, q=0, w=0

Q. For an isothermal free expansion of an ideal gas into vacuum, which one of the following set correct?

$Δ U = 0, q > 0, w < 0$

$Δ U = 0, q > 0, w = 0$

$Δ U = 0, q = 0, w = 0$

$Δ U < 0, q = 0, w < 0$

For isothermal process, $Δ T = 0.$ For an ideal gas, internal energy U depend only on temperature.

So, $Δ U = 0$

Now $Δ U = q + w$

$q = - w = P Δ U$

For free expansion $P_{ext} = 0$

So, $q = - w = 0$

$∴ Δ U = 0, q = 0, w = 0$

Q. For an isothermal free expansion of an ideal gas into vacuum, which one of the following set correct?

ΔU=0,q>0,w<0

ΔU=0,q>0,w=0

ΔU=0,q=0,w=0

ΔU<0,q=0,w<0

For isothermal process, ΔT=0. For an ideal gas, internal energy U depend only on temperature. So, ΔU=0 Now ΔU=q+w q=−w=PΔU For free expansion Pext ​=0 So, q=−w=0 ∴ΔU=0,q=0,w=0

$Δ U = 0, q > 0, w < 0$

$Δ U = 0, q > 0, w = 0$

$Δ U = 0, q = 0, w = 0$

$Δ U < 0, q = 0, w < 0$

For isothermal process, $Δ T = 0.$ For an ideal gas, internal energy U depend only on temperature.

So, $Δ U = 0$

Now $Δ U = q + w$

$q = - w = P Δ U$

For free expansion $P_{ext} = 0$

So, $q = - w = 0$

$∴ Δ U = 0, q = 0, w = 0$