Question

Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K . The enthalpy change (in kJ) for the process is:
a) 11.4 kJ
b) -11.4 kJ
c) 0 kJ
d) 4.8 kJ

Answer 1

Given conditions,

No. of moles(n) = 2 moles.

Temperature(T) = 300 K.

V₂ = 10 L, and V₁ = 1 L.

Universal Gas Constant (R) = 8.31 J/moleK.

Using the formula,

Work done = -2.303nRTlog(V₂/V₁)  [For Chemistry.]

∴ Work done = -2.303 × 2 × 8.31 × 300 × log(10/1)

∴ Work done = -11482.758 J.

∴ Work done = -11.483 kJ.

Now, Process is Isothermal, therefore, Internal Energy will also be zero.

Thus, Using first Law of thermodynamics (of Chemistry and not of physics)

∴ ΔU = ΔQ + w

∴ ΔQ = - w

∴ ΔQ = - (-11.483)

∴  ΔQ = 11.483 kJ ≈ 11.4 kJ.

Now, ΔQ is the enthaly because heat constant at constant pressure is enthalpy.

Hence, Option (a). is true.

Hope it helps.

Answer 2

Answer:

Explanation:

H = E + PV

and ΔH=ΔE+Δ(PV)

or ΔH=ΔE+nRΔT

ΔT=0

ΔE=0

∴ΔH=0