A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ.

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁

   = mRT₁      [∵pV = mRT]

   = (5) (0.287) (272.039)

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU =

From 1st law of thermodynamics

Q = ΔU  + W

   = 0 + 270.588

   = 270.588 kJ

the answer is kevlar

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s