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2 moles of Argon gas is taken in a cylinderfitted with massless and frictionless pistonand occupies a volume of 5 3dm at 298 K.The gas undergoes isothermal reversibleexpansion to 25 3dm . If the gas behavesideally, calculate the molar entropy changefor the process

Question

2 moles of Argon gas is taken in a cylinderfitted with massless and frictionless pistonand occupies a volume of 5 3dm at 298 K.The gas undergoes isothermal reversibleexpansion to 25 3dm . If the gas behavesideally, calculate the molar entropy changefor the process

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Solution

To calculate the molar entropy change for the given process, we can use the formula:

ΔS = nR ln(V2/V1)

Where: ΔS is the molar entropy change n is the number of moles of gas (given as 2 moles) R is the ideal gas constant (8.314 J/(mol·K)) V1 is the initial volume (given as 5 dm^3) V2 is the final volume (given as 25 dm^3)

Let's substitute the given values into the formula:

ΔS = (2 mol) * (8.314 J/(mol·K)) * ln(25 dm^3 / 5 dm^3)

First, let's simplify the volume ratio:

ΔS = (2 mol) * (8.314 J/(mol·K)) * ln(5)

Now, we can calculate the natural logarithm of 5:

ΔS = (2 mol) * (8.314 J/(mol·K)) * 1.609

Finally, we can calculate the molar entropy change:

ΔS ≈ 26.47 J/K

Therefore, the molar entropy change for the given process is approximately 26.47 J/K.

This problem has been solved

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