Question

Remove Looking pin 50 cm 1.0 atm 20°C 100 cm 3.0 am WWW 1. Start 2. Heat to 3 atm. 3. Remove pin. Continue heating to 100 cm 4. Insert pin. Remove mass The figure shows a sample of N2 gas covered by a movable piston, first in its initial state and then undergoing three processes. The heat capacity at constant volume for Nz is 5/2 R. In step 1, when the pin locking the piston is removed, the piston does not move. The piston has an area of 10 cm². The figure gives the volume, pressure, and temperature of the gas. If there is a vacuum above the piston, what is the combined weight of the piston and masses, in Newtons? Please express all answers to three significant figures.

can you help with Q4&Q5 please

Answer 1

Using Ideal Gas equation: PV = nRT

Where P = Pressure, V= Volume, n= no. of moles, R = Universal Gas Constant, T = temperature

Given:

P = 1 atm = 101325 Pa

V = 50 cm³ = 50 * 10^-6 m³

T = 20⁰C = 293 K

PV = nRT

101325 * 50 * 10^-6 = n * 8.314 * 293

n = 2.08 * 10^-3 moles

In step 2, the pin is reinserted so it’s a isochoric process

In isochoric process V = constant

Which implies P/T = constant. It can be written as P₁/T₁ = constant

1*101325/293 = 3*101325/T₂

T₂ = 879 K

Since its an isochoric process, V₂ = 50 cm³

Work done in isochoric process is always zero i.e. = W_isochoric = 0 J

Calculation of amount of heat energy absorbed by the gas

Q = mCp∆T = (n * MW)(5/2 R)(879-293)

Q = (2.08 * 10^-3 X 28)(5/2 * 8.314)(586)

Q = 709.36 J

Therefore the amount of heat energy absorbed by the gas is 709.36 J

Calculation of change in thermal energy of the gas

By first law of thermodynamics: Q – W = ∆E

Q – 0 = ∆E

∆E = 709.36 J

Therefore the change in thermal energy of the gas is 709.36 J