By how much does the entropy of $3$ mole of an ideal gas change in going from a pressure of $2$ bar to a pressure of $1$ bar without any change in temperature. If the surrounding is at $1$ bar and $300$ K. (Expansion is again of the constant external pressure of surrounding).

+ 7.508 J - K^{- 1}

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+ 4.82 J - K^{- 1}

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- 5.29 J - K^{- 1}

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- 8.35 J - K^{- 1}

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Solution

The correct option is A $+ 7.508 J - K^{- 1}$
$Δ S = n R log \frac{P_{1}}{P_{2}}$ For isothermal process
$= 3 \times 8.314 log \frac{2}{1}$
$= 7.508 J / K$ .

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By how much does the entropy of 3 mole of an ideal gas change in going from a pressure of 2 bar to a pressure of 1 bar without any change in temperature. If the surrounding is at 1 bar and 300K. (Expansion is again of the constant external pressure of surrounding).

The correct option is A +7.508 J−K−1ΔS=nRlogP1P2 For isothermal process =3×8.314log21 =7.508J/K .

By how much does the entropy of $3$ mole of an ideal gas change in going from a pressure of $2$ bar to a pressure of $1$ bar without any change in temperature. If the surrounding is at $1$ bar and $300$ K. (Expansion is again of the constant external pressure of surrounding).

The correct option is A $+ 7.508 J - K^{- 1}$
$Δ S = n R log \frac{P_{1}}{P_{2}}$ For isothermal process
$= 3 \times 8.314 log \frac{2}{1}$
$= 7.508 J / K$ .