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:

  • A
    $\frac{3}{2} R \ln 2$
  • B
    $-\frac{3}{2} R \ln 2$
  • C
    $5 R \ln 2$
  • D
    $\frac{5}{2} R \ln 2$

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Similar Questions

For chemical reactions,the calculation of change in entropy is normally done:

At $373 \, K$,steam and water are in equilibrium and $\Delta H = 39.2 \, kJ \, mol^{-1}$. What will be $\Delta S$ for the conversion of $1 \, mole$ of water into steam?
$H_2O_{(l)} \to H_2O_{(g)}$ ... $J \, K^{-1} \, mol^{-1}$

In a spontaneous process,the entropy of the system and its surroundings

$A$ container is divided into two compartments by a removable partition as shown below:
In the first compartment,$n_{1}$ moles of ideal gas $He$ is present in a volume $V_{1}$. In the second compartment,$n_{2}$ moles of ideal gas $Ne$ is present in a volume $V_{2}$. The temperature and pressure in both the compartments are $T$ and $p$,respectively. Assuming $R$ is the gas constant,the total change in entropy upon removing the partition when the gases mix irreversibly is:

For a process to occur spontaneously,

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