An ideal gas, initially at 25oc and 1 bar, undergoes the following cyclic processes in a closed system: (a) in mechanically reversible processes, it is first compressed adiabatically to 5 bar then cooled at a constant pressure of 5 bar to 25oc, and finally expanded isother. mally to its original pressure. (b) the cycle is irreversible and each step has an efficiency of 80% compared with the corresponding mechanically reversible process. the cycle still consists of an adiabatic compression step, an isobaric cooling step, and an isothermal expansion. calculate q, w, au, and ah for each step of the process and for the cycle. take cp (7/2) r and cv (5/2) r.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s