2 moles of ideal gas is expanded isothermally & reversibly from 1 liter to 10 liter. Find the enthalpy change in (kJ mol-1,rounded up to first decimal place ).Correct answer is '0'. Can you explain this answer?

Question

Accepted Answer

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 x 2 x 8.31 x 300 x 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.

Answer

Since the process is isothermal, the temperature remains constant throughout the expansion. Therefore, we can use the ideal gas law to relate the initial and final states of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Since the number of moles and temperature are constant, we can write:

P1V1 = P2V2

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Since we know that there are 2 moles of gas, we can also write:

n = 2

Now, let's assume that the initial pressure and volume are P1 and V1, and the final volume is 2V1. Then we can write:

P1V1 = P2(2V1)

or

P2 = P1/2

This means that the final pressure is half of the initial pressure.

Therefore, the gas expands isothermally from initial conditions of P1 and V1 to final conditions of P1/2 and 2V1.

Answer

I think correct ans is zero ,, because the process is isothermal reversible process ,, so in case of isothermal reversible process DT=0 so DH=0

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