For the isothermal expansion of an ideal gas

E and H increases

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

E increases but H decreases

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

H increases but E decreases

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

E and H are unaltered

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

Open in App

Solution

The correct option is D E and H are unaltered
In isothermal reversible process ideal gas has constant volume and so
$Δ E = 0 a n d Δ H = Δ E = 0$ . This is because both are directly proportional to $Δ T$ which is 0 in this case.

Suggest Corrections

Similar questions

For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ DH = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -

Q. An ideal gas is subjected to a thermodynamics process in which its pressure increases but the volume decreases. Then

Q. The value of the ratio

\frac{C_{p}}{C_{v}}

for

H_{2}

is 1.67 at 30k but decreases to 1.4 at 300K as more degrees of freedom become active. During this rise in temperature,

Q. In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about
(a) 0.25%
(b) 0.5%
(c) 0.7%
(d) 1%.

Q. Assertion :Bond angle of

H_{2} S

is smaller than

H_{2} O

. Reason: Electronegativity of the central atom increases, bond angle decreases.

Join BYJU'S Learning Program

For the isothermal expansion of an ideal gas

The correct option is D E and H are unalteredIn isothermal reversible process ideal gas has constant volume and so ΔE=0 and ΔH=ΔE=0. This is because both are directly proportional to ΔT which is 0 in this case.

The correct option is D E and H are unaltered
In isothermal reversible process ideal gas has constant volume and so
$Δ E = 0 a n d Δ H = Δ E = 0$ . This is because both are directly proportional to $Δ T$ which is 0 in this case.