A frictionless piston-cylinder contains carbon dioxide gas (CO2) initially at 500oC and 2 MPa. The system is o3 cooled in an isobaric process until the final temperature becomes 350 C and the final volume is 1 m . For this process, determine: (a) The reduced pressure and the reduced temperature of the initial state. (b) The initial volume of the piston-cylinder (in m3). (c) The mass of CO2 in the piston-cylinder (in kg). (d) The total boundary work for the process (in kJ). (e) The amount of heat transfer during the cooling process (in kJ).

(a) The reduced pressure is 0.2711 MPa

The reduced temperature is 2.54 K

(b) The initial volume of the piston is approximately 0.806 m³

(c) The mass of CO₂ is approximately 16.9884 kg

(d) The work done, W is approximately 388.023 kJ

(e) The heat transfer is approximately -2,650.1904 kJ

Explanation:

The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K

The initial pressure of the gas, P₁ = 2 MPa

The final temperature of the gas, T₂ = 350°C

The final volume of the gas = 1 m³

(a) For an isobaric process, we have;

The reduced pressure,

The critical pressure of carbon dioxide, = 7.3773 MPa

The reduced pressure, = 0.2711 MPa

The critical temperature, = 304.13 K

The reduced temperature, , is given by the following formula;

Therefore, = (773.15 K)/(304.13 K) = 2.54216947 K

The reduced temperature, ≈ 2.54 K

(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁

∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³  ≈ 0.806 m³

The initial volume of the piston, V₁ ≈ 0.806 m³

(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;

n = P·V/(T·R)

The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles

The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg

(d) The work done, W = P·()

W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227

The work done, W ≈ 388.023 kJ

(e) The heat transfer dQ = m· ×(T₂ - T₁)

for CO₂ ≈ 1.04 kJ/(kg·K)

∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ

Therefore, the heat transfer = dQ = -2,650.1904 kJ

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: