Calculate the increase in entropy (approximat | vedclass

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(A) The change in entropy for an ideal gas is given by the formula: $\Delta S = n C_p \ln(\frac{T_2}{T_1}) - nR \ln(\frac{P_2}{P_1})$.Given: $n = 3 \

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) The change in entropy for an ideal gas is given by the formula: $\Delta S = n C_p \ln(\frac{T_2}{T_1}) - nR \ln(\frac{P_2}{P_1})$.Given: $n = 3 \ mol$,$C_p = 7 \ cal/mol \cdot K$,$T_1 = 300 \ K$,$T_2 = 1000 \ K$,$P_1 = 0.1 \ atm$,$P_2 = 1 \ atm$,$R \approx 2 \ cal/mol \cdot K$.Substituting the values:$\Delta S = 3 	imes 7 	imes 2.3 \log(\frac{1000}{300}) - 3 	imes 2 	imes 2.3 \log(\frac{1}{0.1})$$\Delta S = 21 	imes 2.3 \log(3.33) - 6 	imes 2.3 \log(10)$Using $\log(3.33) \approx 0.52$ and $\log(10) = 1$:$\Delta S = 48.3 	imes 0.52 - 13.8 	imes 1$$\Delta S = 25.116 - 13.8 = 11.316 \ cal/K$.The approximate value is $11 \ cal/degree$.

Calculate the increase in entropy (approximate) of $3 \ mol$ of hydrogen gas as it changes from $300 \ K$ at $0.1 \ atm$ to $1000 \ K$ and $1 \ atm$ $(C_p = 7 \ cal/degree/mol)$. $[log \ 3 = 0.48, \ ln \ x = 2.3 \ log \ x]$ .......$cal/degree$

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