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Assuming that the volumes are additive, and the gases behave ideally, calculate the entropy of 12.0 L of N2 with 4.00 L of N2O at 250 K and 0.550 atm. (20 points) Write a meaningful comment on your findings. (5 points)

Respuesta :

This problem is describing a gaseous system composed by 12.0 L of nitrogen and 4.00 L of dinitrogen monoxide at 250 K and 0.550 atm. In addition, it is asking for the entropy of the system assuming it is ideal and the volumes are additive; this means we can use the following equation for the entropy of the mixture:

[tex]S_{mix}=-n_{N_2}*R*ln(\frac{n_{N_2}}{n_{N_2}+n_{N_2O}} )-n_{N_2O}*R*ln(\frac{n_{N_2O}}{n_{N_2}+n_{N_2O}} )[/tex]

It means we need to calculate the moles of both gases, which can be obtained by using the ideal gas equation for the common T,P conditions and the volume of each gas:

[tex]n_{N_2}=\frac{PV_{N_2}}{RT} =\frac{0.550atm*12.0L}{0.08206\frac{atm*L}{mol*K}*250K} =0.322mol\\\\n_{N_2O}=\frac{PV_{N_2O}}{RT} =\frac{0.550atm*4.00L}{0.08206\frac{atm*L}{mol*K}*250K} =0.107mol[/tex]

Next, we recall the aforementioned equation for the calculation of the entropy and plug in the values:

[tex]S_{mix}=-8.3145\frac{J}{mol*K}[0.322mol*ln(\frac{0.322mol}{0.322mol+0.107mol} )+0.107mol*ln(\frac{0.107mol}{0.322mol+0.107mol} )]\\\\S_{mix}=2.00\frac{J}{mol*K}[/tex]

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