When 2 mol of an ideal gas left( C_{p,m} = f | vedclass

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(C) The change in entropy $\left( \Delta S \right)$ for an ideal gas at constant pressure is given by the formula: $\Delta S = n C_{p,m} \ln \left( \f

Vedclass · Accepted Answer

$5 R \ln 2$

Vedclass · Answer

(C) The change in entropy $\left( \Delta S ight)$ for an ideal gas at constant pressure is given by the formula: $\Delta S = n C_{p,m} \ln \left( \frac{T_2}{T_1} ight)$.Given values are: $n = 2 \ mol$,$C_{p,m} = \frac{5}{2} R$,$T_1 = 300 \ K$,and $T_2 = 600 \ K$.Substituting these values into the formula:$\Delta S = 2 	imes \left( \frac{5}{2} R ight) \ln \left( \frac{600 \ K}{300 \ K} ight)$.$\Delta S = 5 R \ln (2)$.Therefore,the correct option is $C$.

When $2 \ mol$ of an ideal gas $\left( C_{p,m} = \frac{5}{2} R \right)$ is heated from $300 \ K$ to $600 \ K$ at constant pressure,the change in entropy of the gas $\left( \Delta S \right)$ is:

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