these all questions are digrammatic......

1.) figure shows a cylindrical tube of radius 5 cm and lenght 20 cm. it is closed by a tight-fitting cork. the friction coefficient b/w the cork and the tube is 0.2 . the tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K.the tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. let dN denote the magnitude of the normal contact force exerted by a small lenght dl of the cork along the periphery..assuming that the temeprature of the gas is unform at any instant, calculate dN/dl

2.figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas . the internal energy of this ideal gas is given by 1.5nRT.the tube is divided into two equal parts by a fixed diathermic walls initially, the pressure & the temperature are as shown.the system is left for sufficient time so that the temperature become equal on the two sides.

a) find the final pressures on the two sides.

b) find the final equilibrium temperature

c) how much heat has flown from the fgas on the right to the gas on the left??

3.calculate the heat absorbed ny a system in going through the cyclic process as shown...

Let's break down your questions one by one, focusing on each scenario involving thermodynamics and gas laws. We'll start with the first question regarding the cylindrical tube and the cork.

1. Normal Contact Force in a Heated Tube

In this scenario, we have a cylindrical tube with a cork that pops out when the temperature reaches 600 K. The friction coefficient between the cork and the tube is given as 0.2. To find the magnitude of the normal contact force per unit length, denoted as dN/dl, we can use the relationship between pressure, area, and force.

Understanding the Forces at Play

The pressure inside the tube is initially 1 atm, which we can convert to pascals for calculations:

• 1 atm = 101325 Pa

The force exerted by the gas on the cork can be calculated using the formula:

F = P × A

Where:

• P is the pressure of the gas (in pascals).
• A is the area of the cork in contact with the gas.

The area A for the cork can be calculated as:

A = 2πr × dl

Here, r is the radius of the tube (5 cm = 0.05 m), and dl is the small length of the cork we are considering. Thus, the force exerted by the gas on this small length of cork becomes:

F = P × (2πr × dl)

Now, the normal contact force dN exerted by the cork on the tube can be expressed as:

dN = μ × F

Where μ is the coefficient of friction (0.2). Substituting for F, we have:

dN = μ × P × (2πr × dl)

To find dN/dl, we divide both sides by dl:

dN/dl = μ × P × 2πr

Substituting the values:

• μ = 0.2
• P = 101325 Pa
• r = 0.05 m

Calculating dN/dl:

dN/dl = 0.2 × 101325 × 2π × 0.05

After performing the calculation, you will find the value of dN/dl.

2. Equilibrium in a Diathermic Wall System

Now, let's move on to the second question regarding the two equal parts of the tube divided by a diathermic wall. We need to find the final pressures, equilibrium temperature, and the heat transfer between the gases.

Final Pressures and Equilibrium Temperature

Initially, let’s denote the pressures and temperatures of the two sides as P1, T1 and P2, T2. Since the walls are diathermic, heat can flow between the two sides until thermal equilibrium is reached.

Using the ideal gas law, we can express the pressures in terms of the number of moles (n), volume (V), and temperature (T):

P = nRT/V

Since the volumes are equal and the gases will reach the same final temperature (Tf), we can set up the equations:

• P1 = n1RT1/V
• P2 = n2RT2/V

At equilibrium, the total number of moles and the total energy must be conserved. Therefore, we can use the conservation of energy to find Tf:

n1C_vT1 + n2C_vT2 = (n1 + n2)C_vTf

From this equation, we can solve for Tf:

Tf = (n1T1 + n2T2) / (n1 + n2)

Once we have Tf, we can substitute back to find the final pressures using the ideal gas law.

Heat Transfer Calculation

The heat transferred from the gas on the right to the gas on the left can be calculated using:

Q = nC_v(Tf - T2)

Where n is the number of moles on the right side, C_v is the specific heat at constant volume, and Tf is the final equilibrium temperature. This will give you the amount of heat that has flowed from the gas on the right to the gas on the left.

3. Heat Absorbed in a Cyclic Process

For the cyclic process, we need to analyze the specific path taken by the system on a PV diagram. The heat absorbed during a cyclic process can be calculated by considering the area enclosed by the cycle on the PV diagram, which represents the net work done on or by the system.

To calculate the heat absorbed, you can use:

Q = W + ΔU

Where W is the work done by the system, and ΔU is the change in internal energy. In a complete cycle, ΔU is zero, so:

Q = W

By analyzing the specific segments of the cycle (isothermal, adiabatic, etc.), you can calculate the work done in each segment and sum them up to find the total heat absorbed by the system.

Each of these problems involves applying fundamental principles of thermodynamics, and with careful calculations, you can arrive at the correct answers. If you have any specific values or additional details for these problems, feel free to share, and we can dive deeper into the calculations together!