A 100-L insulated tank contains N2 gas at 200 kPa and ambient temperature 25^∘ C. The tank is co

A 100-L insulated tank contains N2 gas at 200 kPa and...

$\mathrm{A} 100-\mathrm{L}$ insulated tank contains $\mathrm{N}_{2}$ gas at 200 $\mathrm{kPa}$ and ambient temperature $25^{\circ} \mathrm{C}$. The tank is connected by a valve to a supply line flowing $\mathrm{CO}_{2}$ at $1.2 \mathrm{MPa}, 90^{\circ} \mathrm{C} .$ A mixture of $50 \% \mathrm{N}_{2}$ $50 \% \mathrm{CO}_{2}$ by mole should be obtained by opening the valve and allowing $\mathrm{CO}_{2}$ to flow in until an appropriate pressure is reached, when the valve is closed. What is the pressure? The tank eventually cools to ambient temperature. Find the net entropy change for the overall process.

$\mathrm{A} 100-\mathrm{L}$ insulated tank contains $\mathrm{N}_{2}$ gas at 200 $\mathrm{kPa}$ and ambient temperature $25^{\circ} \mathrm{C}$. The tank is connected by a valve to a supply line flowing $\mathrm{CO}_{2}$ at $1.2 \mathrm{MPa}, 90^{\circ} \mathrm{C} .$ A mixture of $50 \% \mathrm{N}_{2}$
$50 \% \mathrm{CO}_{2}$ by mole should be obtained by opening the valve and allowing $\mathrm{CO}_{2}$ to flow in until an appropriate pressure is reached, when the valve is closed. What is the pressure? The tank eventually cools to ambient temperature. Find the net entropy change for the overall process.

Key Concepts

Ideal Gas Law

The ideal gas law, PV = nRT, is fundamental in thermodynamics and is used to relate the pressure, volume, temperature, and amount of an ideal gas. This concept allows one to calculate the number of moles present in a given volume at specified conditions, which is essential when analyzing processes involving gas mixing and changes in state.

Mole Balance and Gas Mixing

In processes where different gases are combined, the mole balance is a critical concept. It involves accounting for the individual contributions of each gas to the total number of moles present in the mixture, ensuring that the target mole fractions are met. This concept underpins the determination of how much of one gas must be added to achieve a desired composition in the final mixture.

Entropy Change in Mixing

The change in entropy accompanying the mixing of gases arises from the increase in disorder when individual molecules that were previously separated become intermingled. The entropy of mixing is calculated using the mole fractions of the components and reflects the irreversible nature of the mixing process, contributing a net positive entropy change.

Irreversible Process Analysis

Many practical mixing operations are inherently irreversible, meaning that even though the final state functions (such as pressure, temperature, and composition) are well defined, the path taken to reach them involves non-equilibrium states. When evaluating such processes, one can compute the net entropy change by comparing the initial and final equilibrium states, as entropy is a state function.

Adiabatic and Isothermal Processes

An adiabatic process involves no heat exchange with the surroundings, while an isothermal process maintains a constant temperature. In some mixing scenarios, a system may initially undergo an adiabatic process followed by heat exchange that brings it to thermal equilibrium at a desired temperature. Understanding these processes is crucial when analyzing overall changes in state and entropy in thermodynamic systems.

Transcript

00:01 Hi, everybody.

00:01 So this question is asking us, find the net intrepid change with the overall process, and what is the pressure, okay, inside the tank.

00:14 So from this, let me get to, we need to find the number of moles in the tank.

00:25 So we have n2 co2 is going to equal to n2 in 2, n2, n2, n equal to 1, n2, okay? and we have p1, pressure 1, and this is all for the first state, okay? and rt1 equals 200 times 0 .1 divided by 8 .3145 times 273 .15 equals 25 equals 0 .15 equals 0 .15 equals 0 .008030 .0 .3 .3.

01:21 6, 7, 8 kilomoles.

01:27 Okay.

01:28 And the total number of moles in the tank at state 2 is we have n2 equals n2n2 or nitrogen 2.

01:42 There we go.

01:44 Plus n2 co2.

01:48 And this equals to 0 .080.

01:54 678 plus 0 .008678 equals 0 .01165 kilo mules.

02:09 Okay.

02:12 And now we've got to fly the first law to the system.

02:17 So we get the first law.

02:20 And so we have n initial h, initial plus.

02:26 N1 equals n2 u2.

02:31 And now we have n initial co2, cp initial times temperature initial.

02:40 Make that t look nice, that's a weird t, plus n1, cv1, t1n2 equals, i think it's that equal looking nicer, equals an initial co2c, cv initial co2, 2 plus n1n plus n1n2, cv1 in 2 times t2.

03:20 Okay, and here are n1, 1, 1, 1, initial, and so because of that, we have cp initial, t initial, plus cv1, t1 equals cv initial plus cv1 times t2.

03:48 And now we have, let's see, there go.

04:08 So now we have 0 .842 times 4 .1 times 90 plus 273 .15.

04:22 So we're just trying to convert it to caldons plus 0 .745 times 28 .013 times 298 .15 equals 298 .15 equals 2 .98 .15 equals 2 .5.

04:38 0 .653.

04:45 And what i'm going to do is there we go times 4 .01 plus 0 .745 times 28 .13 times t2.

05:08 And so we're going to keep going so we have 1 .3 457 .03 plus 622 .2 .29.

05:24 That's going to make sure i add enough to use there.

05:27 Equals 49 .68 -t2.

05:32 So your t2 is going to equal to 396 .69.

05:39 Kelvin's and we can find the final pressure of the tank is p2 equals into rt2 v and so we got 0 .016135 times 8 .3145 times 396 .6.

06:04 Divided by 0 .1 equals 532 .177.

06:14 Okay.