Two rigid, insulated tanks are connected with a pipe and...

Two rigid, insulated tanks are connected with a pipe and valve. One tank has $0.5 \mathrm{~kg}$ air at $200 \mathrm{kPa}, 300 \mathrm{~K}$ and the other has $0.75 \mathrm{~kg}$ air at $100 \mathrm{kPa}, 400 \mathrm{~K}$. The valve is opened, and the air comes to a single uniform state without any heat transfer. Find the final temperature and the change in entropy of the air.

Key Concepts

Entropy Change in Irreversible Processes

Entropy quantifies the degree of disorder in a system and increases in spontaneous, irreversible processes. When different gas states mix, the irreversible nature of the process leads to entropy generation, which can be computed by considering the entropy changes of the individual gas portions even in the absence of heat transfer.

Ideal Gas Behavior

Ideal gases follow the simple relationship given by the ideal gas law, which links pressure, volume, temperature, and mass. This model is particularly useful for air, enabling the calculation of state variables and simplification of energy and entropy analyses during mixing and other thermodynamic processes.

Energy Conservation (First Law of Thermodynamics)

The principle of energy conservation states that the total energy of an isolated system remains constant. For a closed system undergoing a mixing process without heat transfer, the sum of the internal energies before and after the process must be equal, allowing determination of the final equilibrium state such as the uniform temperature.

Adiabatic Process

An adiabatic process is one in which there is no heat transfer between the system and its surroundings. In insulated systems, any thermodynamic changes occur without the exchange of energy as heat, meaning that the internal energy changes are solely due to work interactions or, in the case of mixing, redistribution of energy among components.

Transcript

00:01 In this question it is given that there are two rigid and insulated tanks.

00:05 One have the air of mass 0 .5 kilogram.

00:12 Its initial pressure is 200 kylpascal and the temperature is 300 kelvin and the other tank contains 0 .75 kilogram of air.

00:26 The pressure is 100 kylpascal and the temperature is 400 kelvin.

00:35 These two tanks are connected with the help of a pipe and valve system and initially the valve is closed.

00:45 Now the valve is opened and air is allowed to mix so the complete air comes into a single uniform state without any heat transfer.

00:57 We are required to evaluate the final temperature of air and the change in entropy of the air.

01:03 So let's see how to solve this question.

01:05 First of all, let's apply the formula to calculate volume of air in the first tank or we can say in the tank a.

01:17 So we will have volume a is equal to m -a -r, p -a -1 divided by p -a -1.

01:29 Now substitute all the values.

01:31 So we will have v -a is equal to m -a.

01:35 That means we are talking about mass of the air in the first tank.

01:40 So that is equals to 0 .5 multiplied by r, that means universal gas constant for the air, and we know that it is equals to 0 .287 multiplied by temperature, and the temperature of the air in the first tank is 300 kelvin, and it is divided by pa1, that means 200 kioskl.

02:03 So when we further calculate, we get volume of air in tank a or in the first tank va is equal.

02:11 To 0 .2153 meter cube.

02:17 Similarly again apply the same formula to calculate volume of air in the second tank or the tank b so we can write vb is equal to mb r p b1 divided by p b1 now substitute all the values so we will have vb is equal to 0 .0 0 .75 multiplied by 0 .287 multiplied by 400 upon 100 and when we further calculate we get vb is equal to 0 .861 meter cube.

03:01 So now we have the volumes of the air in different tanks.

03:06 Now let's find the total mass of air.

03:15 So we can write total mass or we are denoting it as m2 and this will be equals to m a mass of air in the tank a or in the first tank plus mb or mass of air in tank b or in the second tank...

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