Question

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide, 25% oxygen, 35% neon The cylinder contains 0.8 kg of the mixture at 260 oC and 450 kPa. Determine the magnitude and direction of work when the mixture undergoes an isobaric process to 95 oC.

Answer 1

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.