(a). A vertical cylinder of height 100cm contains air at a constant temp.
The top is closed by a frictionless piston . The atm. pressure is equal to75 cm of mercury . Mercury is slowly pour over piston.Find maximum height of mercury column that can be put on the piston.

(b)A container of volume 50cc contains air (mean molecular weight =28.8gm)and is open to atmosphere where the pressure is 100kPa
The container is kept in bath containing melting ice(0 degree)
1.find mass of the air in the container when thermal equilibrium is reached
2. the cotainer is now placed in another bath containing boiling (100degree)find mass of air in the container

Let's break down your questions step by step, starting with part (a) regarding the vertical cylinder with a piston and mercury. We need to determine the maximum height of the mercury column that can be placed on the piston while maintaining equilibrium.

Understanding the Cylinder and Piston System

The cylinder is closed at the top by a frictionless piston, and we know the atmospheric pressure is equivalent to 75 cm of mercury. To find the maximum height of the mercury column, we can use the principles of fluid statics and pressure equilibrium.

Pressure Equilibrium

In this scenario, the pressure exerted by the mercury column must balance the pressure inside the cylinder plus the atmospheric pressure. The pressure exerted by a column of mercury can be calculated using the formula:

• P = h * ρ * g

Where:

• P is the pressure (in pascals),
• h is the height of the mercury column (in meters),
• ρ is the density of mercury (approximately 13,600 kg/m³), and
• g is the acceleration due to gravity (approximately 9.81 m/s²).

Calculating the Maximum Height

The pressure inside the cylinder can be expressed as:

• P_inside = P_atm + P_mercury

Given that the atmospheric pressure is 75 cm of mercury, we convert this to meters:

• P_atm = 0.75 m * 13,600 kg/m³ * 9.81 m/s² ≈ 10,000 Pa

Now, we can set up the equation:

• P_mercury = h * ρ * g

Substituting the values:

• 10,000 Pa = h * 13,600 kg/m³ * 9.81 m/s²

Solving for h gives:

• h ≈ 0.075 m or 7.5 cm

This means the maximum height of the mercury column that can be placed on the piston is approximately 7.5 cm.

Analyzing the Air in the Container

Now, let's move on to part (b), where we have a container of 50 cc containing air at a pressure of 100 kPa and a mean molecular weight of 28.8 g/mol. We need to find the mass of air in the container at two different temperatures.

Finding Mass at 0 Degrees Celsius

To find the mass of air in the container when thermal equilibrium is reached at 0 degrees Celsius, we can use the ideal gas law:

• PV = nRT

Where:

• P is the pressure (in pascals),
• V is the volume (in cubic meters),
• n is the number of moles of gas,
• R is the universal gas constant (8.314 J/(mol·K)), and
• T is the temperature (in Kelvin).

First, convert the volume from cc to cubic meters:

• V = 50 cc = 50 x 10^-6 m³

Next, convert the temperature to Kelvin:

• T = 0 + 273.15 = 273.15 K

Now, we can rearrange the ideal gas law to find the number of moles:

• n = PV / RT

Substituting the values:

• n = (100,000 Pa) * (50 x 10^-6 m³) / (8.314 J/(mol·K) * 273.15 K)

Calculating this gives:

• n ≈ 0.0022 moles

To find the mass, we multiply the number of moles by the mean molecular weight:

• mass = n * M

Where M is the mean molecular weight in kg/mol:

• mass = 0.0022 moles * 0.0288 kg/mol ≈ 0.00006336 kg or 0.06336 g

Finding Mass at 100 Degrees Celsius

Now, let's calculate the mass of air when the container is placed in a bath at 100 degrees Celsius. Convert the temperature to Kelvin:

• T = 100 + 273.15 = 373.15 K

Using the ideal gas law again:

• n = PV / RT

Substituting the new temperature:

• n = (100,000 Pa) * (50 x 10^-6 m³) / (8.314 J/(mol·K) * 373.15 K)

Calculating this gives:

• n ≈ 0.0016 moles

Now, finding the mass:

• mass = 0.0016 moles * 0.0288 kg/mol ≈ 0.00004608 kg or 0.04608 g

In summary, the mass of air in the container at 0 degrees Celsius is approximately 0.06336 g, and at 100 degrees Celsius, it is approximately 0.04608 g