There are 1.36 kgm of air at 137.9 Kpaa stirred with internal paddles in an insulated rigid container whose volume is 0.142 m3, until the pressure becomes 689.5 Kpaa. Determine (a) the work input, (b) Δ(pV), and (c) Q. Cp=1.0062 kj/(Kgm.°K), Cv = 0.7186 kj/(Kgm.°K), R= 287.18 j/(Kgm.°K)

To determine the answers to the given problem, we need to use the laws of thermodynamics, specifically the ideal gas law and the first law of thermodynamics.

(a) To calculate the work input (W), we can use the equation:

W = ΔU - Q

where ΔU is the change in internal energy and Q is the heat transferred.

First, let's calculate the change in internal energy (ΔU) using the equation:

ΔU = m * Cv * ΔT

where m is the mass of air, Cv is the specific heat at constant volume, and ΔT is the change in temperature.

Given:
Mass of air (m) = 1.36 kg
Specific heat at constant volume (Cv) = 0.7186 kJ/(kg·K)
Change in temperature (ΔT) = Tf - Ti = 137.9 Kpaa - 689.5 Kpaa = -551.6 Kpaa

Note: Kpaa represents kilopascal absolute, which is the pressure relative to absolute zero (0 K).

Now, we can substitute the given values into the equation:

ΔU = (1.36 kg) * (0.7186 kJ/(kg·K)) * (-551.6 Kpaa)

Simplifying the units:

ΔU = (1.36 kg) * (0.7186 kJ/(kg·K)) * (-551.6 Kpaa) * (1000 J/kJ)

ΔU = -528,910.464 J

Next, we need to calculate the heat transferred (Q). Since the process is adiabatic (insulated), no heat is transferred:

Q = 0 J

Finally, we can calculate the work input (W):

W = ΔU - Q
W = -528,910.464 J - 0 J
W = -528,910.464 J

The work input is approximately -528,910.464 J.

(b) To calculate Δ(pV), we can use the ideal gas law:

pV = mRT

where p is the pressure, V is the volume, m is the mass of air, R is the specific gas constant.

Given:
Initial pressure (p1) = 137.9 Kpaa
Final pressure (p2) = 689.5 Kpaa
Initial volume (V1) = 0.142 m^3
Final volume (V2) = 0.142 m^3
Mass of air (m) = 1.36 kg
Specific gas constant (R) = 287.18 J/(kg·K)

We can rearrange the ideal gas law equation to calculate Δ(pV):

Δ(pV) = (p2 * V2 - p1 * V1)

Substituting the given values into the equation:

Δ(pV) = (689.5 Kpaa * 0.142 m^3 - 137.9 Kpaa * 0.142 m^3)

Simplifying the units:

Δ(pV) = (689.5 Kpaa - 137.9 Kpaa) * 0.142 m^3

Δ(pV) = 77.19 Kpa·m^3

Therefore, Δ(pV) is approximately 77.19 Kpa·m^3.

(c) To calculate Q (heat transfer), we can use the first law of thermodynamics:

Q = ΔU - W

Given:
ΔU = -528,910.464 J (from part a)
W = -528,910.464 J (from part a)

Substituting the given values into the equation:

Q = -528,910.464 J - (-528,910.464 J)

Simplifying:

Q = 0 J

Therefore, Q (heat transfer) is approximately 0 J, as this is an adiabatic process (insulated container).

To summarize:
(a) The work input is approximately -528,910.464 J.
(b) Δ(pV) is approximately 77.19 Kpa·m^3.
(c) Q (heat transfer) is approximately 0 J.