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The temperature of 2.00 mol Ne(g) is increased from 25°C to 200°C at constant pressure. Assuming the heat capacity of Ne is 20.8 J/K·mol, calculate the change in the entropy of neon. Assume ideal gas behavior.a.+19.2 J/Kb.-7.68 J/Kc.+7.68 J/Kd.-19.2 J/K

Respuesta :

Answer:

a) + 19.2J/K

Explanation:

The parameters in this question are outlined as follows;

Number of moles (n)= 2

Initial temperature (T1) = 25°C + 273 = 298 K (Converting to Kelvin)

Final temperature (T2) = 200°C + 273 = 473 K (Converting to Kelvin)

Heat Capacity (Cp) = 20.8 J/K·mol

Change in entropy (ΔS) ?

The formular linking the parameters is given as;

ΔS =n * Cp * ln(T2/T1 )  (constant pressure)

Inserting the values into the equation, we have;

ΔS = 2 * 20.8 * ln (473/298)

ΔS = 41.6 * 0.462

ΔS = 19.2J/K

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