two moles of an ideal gas is expanded isothrmally and reversibly from

1.	two moles of an ideal gas is expanded isothrmally and reversibly from 1 to 10 litre at 300k the enthalphy change is
Answer» Answer: Explanation: 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 Read more on Brainly.in - brainly.in/question/7911313#readmore

two moles of an ideal gas is expanded isothrmally and reversibly from 1 to 10 litre at 300k the enthalphy change is

Answer»

Answer:

Explanation:

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

Read more on Brainly.in - brainly.in/question/7911313#readmore

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