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A particular container holds 4.45 mol of neon gas. The volume of this container can be altered by sliding a piston in or out. The volume is changed from 7.60 L to 1.70 L while at the same time the temperature is changed from 335 K to 3.00 × 102 K. The molar heat capacity, CV,m, for neon is 12.47 J/(mol·K). Assume that this value will not change over the given temperature range. What is the change in entropy for the gas?

Delta S=__________J/K

Respuesta :

Answer : The  change in entropy for the gas is -61.52 J/K

Explanation :

Formula used :

[tex]\Delta S=nC_v\ln (\frac{T_2}{T_1})+nR\ln (\frac{V_2}{V_1})[/tex]

where,

[tex]\Delta S[/tex] = change in entropy  = ?

n = number of moles of gas = 4.45 mole

R = gas constant  = 8.314 J/mol.K

[tex]V_1[/tex] = initial volume of gas  = 7.60 L

[tex]V_2[/tex] = final volume of gas  = 1.70 L

[tex]T_1[/tex] = initial temperature of gas  = 335 K

[tex]T_2[/tex] = final temperature of gas  = [tex]3.00\times 10^2K[/tex]

[tex]C_v[/tex] = molar heat capacity for neon gas = 12.47 J/mol.K

Now put all the given values in the above formula, we get:

[tex]\Delta S=(4.45mol)\times (12.47J/mol.K)\ln (\frac{3.00\times 10^2K}{335K})+(4.45mole)\times (8.314J/mol.K)\ln (\frac{1.70L}{7.60L})[/tex]

[tex]\Delta S=-61.52J/K[/tex]

Therefore, the change in entropy for the gas is -61.52 J/K

onyebuchinnaji

The change in entropy for the neon gas is - 61.53 J/K.

The given parameters;

  • number of moles of the given gas, n = 4.45 moles
  • initial volume of the gas, V₁ = 7.6 L
  • final volume of the gas, V₂ = 1.7 L
  • initial temperature of the gas, T₁ = 335 K
  • final temperature of the gas, T₂ = 300 K
  • molar heat capacity of the gas, Cv =  12.47 J/(mol·K)

The change in entropy for the neon gas is calculated as follows;

[tex]\Delta S = nRln(\frac{V_2}{V_1} ) \ + \ nC_v ln (\frac{T_2}{T_1})\\\\[/tex]

where;

  • R is the ideal gas constant = 8.314 J/mol.K

Substitute the given parameters and solve for the entropy change;

[tex]\Delta S = (4.45)(8.314)ln (\frac{1.7}{7.6} ) \ + \ (4.45)(12.47) ln (\frac{300}{335} )\\\\\Delta S = -61.53 \ J/K[/tex]

Thus, the change in entropy for the neon gas is - 61.53 J/K.

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