For which of the following processes,Delta S | vedclass

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(B) . The conversion of diamond to graphite involves a structural change where entropy generally increases,so $\Delta S > 0$.$B$. When the pressure of

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$N_2(g, 1 \ atm) 	o N_2(g, 5 \ atm)$

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(B) . The conversion of diamond to graphite involves a structural change where entropy generally increases,so $\Delta S > 0$.$B$. When the pressure of a gas increases at constant temperature,the gas molecules are compressed into a smaller volume,leading to a decrease in the number of available microstates and a decrease in randomness. Thus,$\Delta S $C$. Increasing the temperature of a gas increases the kinetic energy and the randomness of the molecules,leading to an increase in entropy,so $\Delta S > 0$.$D$. The dissociation of $H_2$ gas into $2H$ atoms increases the number of moles of gas particles,which increases the disorder of the system,so $\Delta S > 0$.

For which of the following processes,$\Delta S$ is negative?

Similar Questions

Observe the following reactions:
$I$. $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$
$II$. $Cl_{2(g)} \rightarrow 2 Cl_{(g)}$
$III$. $H_2O_{(l)} \rightarrow H_2O_{(s)}$
Identify the reactions in which entropy increases.

What is the entropy of vaporisation of water at $100\,^{\circ}C$,if the molar heat of vaporisation is $9710\,cal/mol$?

Standard entropy of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \ J \ K^{-1} \ mol^{-1},$ respectively. For the reaction,$\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3, \ \Delta H = -30 \ kJ,$ to be at equilibrium,the temperature will be ............... $K$.

Assertion : Water in liquid state is more stable than ice at room temperature.
Reason : Water in liquid form has higher entropy than ice.

$9.0 \, g$ of $H_2O$ is vaporized at $100 \, ^\circ C$ and $1 \, atm$ pressure. If the latent heat of vaporization of water is $x \, J / g$,then $\Delta S$ is given by :-

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