Respuesta :
Explanation:
The given data is as follows.
[tex]T_{1}[/tex] = (19 + 273) K = 292 K, [tex]P_{1}[/tex] = 1.0 atm,
[tex]V_{1}[/tex] = 1.4 L
Therefore, moles of gas present will be calculated as follows.
n = [tex]\frac{P_{1}V_{1}}{RT_{1}}[/tex]
= [tex]\frac{1 atm \times 1.4}{0.0821 L atm/mol K \times 292}[/tex]
= [tex]\frac{1.4}{23.97}[/tex]
= 0.058 mol
Since, neon is a monoatomic ideal gas hence, expression for its constant volume heat capacity is as follows.
[tex]C_{v} = \frac{3}{2}R[/tex]
As the gas is expanding to a volume ([tex]V_{2}[/tex]) of 2.8 L and it is heated to [tex]35^{o}C[/tex] or (35 + 273) K = 308 K.
So, entropy change for this reaction will be calculated as follows.
[tex]\Delta S = nC_{v} ln (\frac{T_{2}}{T_{1}}) + nR ln (\frac{V_{2}}{V_{1}})[/tex]
= [tex]n \times \frac{3}{2}R \times ln (\frac{T_{2}}{T_{1}}) + nR ln (\frac{V_{2}}{V_{1}})[/tex]
= [tex]nR[1.5 ln (\frac{T_{2}}{T_{1}}) + ln (\frac{V_{2}}{V_{1}})[/tex]
= [tex]0.058 \times 8.314 J/mol [1.5 ln (\frac{308}{292}) + ln (\frac{2.8}{1.4})[/tex]
= [tex]0.4822 \times [1.5 \times 0.052 + 0.693][/tex]
= 0.372 J/K
Thus, we can conclude that the entropy change for the process is 0.372 J/K.
