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 work input (a), Δ(pV) (b), and Q (c), we can use the following formulas:
(a) Work input (W) = ΔU - Q
(b) Δ(pV) = W
(c) Q = ΔU - W
Here's how we can calculate each value step-by-step:
Step 1: Calculate the initial internal energy (U₁):
Given that there are 1.36 kgm of air, Cp = 1.0062 kj/(Kgm.°K), and the initial temperature (T₁) = 137.9 K:
U₁ = m * Cp * T₁
U₁ = 1.36 kgm * 1.0062 kj/(Kgm.°K) * 137.9 K
U₁ = 187.524 kj
Step 2: Calculate the final internal energy (U₂):
Given that Cp = 1.0062 kj/(Kgm.°K), Cv = 0.7186 kj/(Kgm.°K), R = 287.18 j/(Kgm.°K), and the final pressure (P₂) = 689.5 Kpaa:
First, convert pressure from Kpaa to Pa:
P₂ = 689.5 Kpaa * 1000 Pa/Kpaa
P₂ = 689,500 Pa
Next, calculate the final temperature (T₂) using the ideal gas law:
P₁ * V₁ / T₁ = P₂ * V₂ / T₂
Given that V₁ = V₂ (insulated rigid container), rearrange the equation:
T₂ = (P₁ * T₁ * V₂) / (P₂ * V₁)
T₂ = (137.9 K * 0.142 m³) / 689,500 Pa
T₂ ≈ 0.0284 K
Now calculate the final internal energy:
U₂ = m * Cv * T₂
U₂ = 1.36 kgm * 0.7186 kj/(Kgm.°K) * 0.0284 K
U₂ ≈ 0.0284 kj
Step 3: Calculate the work input (W):
W = U₂ - U₁
W = 0.0284 kj - 187.524 kj
W ≈ -187.495 kj
Step 4: Calculate Δ(pV):
Given that Δ(pV) = W:
Δ(pV) = -187.495 kj
Step 5: Calculate Q:
Q = U₂ - W
Q ≈ 0.0284 kj - (-187.495 kj)
Q ≈ 187.523 kj
Therefore, the results are:
(a) The work input (W) ≈ -187.495 kj
(b) Δ(pV) ≈ -187.495 kj
(c) The heat transfer (Q) ≈ 187.523 kj
To find the answers to the questions, we'll need to use different formulas and equations related to work, change in pressure and volume, and heat transfer.
(a) To find the work input (W), we'll use the formula:
W = -ΔU
Where ΔU is the change in internal energy of the system.
To calculate ΔU, we need to use the following equation:
ΔU = Q - W
Where Q is the heat transferred to the system.
(b) To find Δ(pV), we'll use the formula:
Δ(pV) = p2V2 - p1V1
Where p1 and p2 are the initial and final pressures, and V1 and V2 are the initial and final volumes.
(c) To find Q, we'll use the equation:
Q = ΔU + W
Now let's calculate each part step by step.
(a) Work Input (W):
Using the formula W = -ΔU, we need to calculate the change in internal energy ΔU.
ΔU = Q - W
As the problem states that the container is insulated, there is no heat transfer (Q = 0) into or out of the system. Therefore, the equation simplifies to:
ΔU = -W
Now, we need to determine the change in internal energy ΔU. Since the container is rigid, there is no change in volume, so the change in internal energy is only due to the change in temperature.
ΔU = m * (Cv) * ΔT
Where m is the mass of air, Cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.
Given:
m = 1.36 kg
Cv = 0.7186 kJ/(kg·K)
ΔT = T2 - T1 = (689.5 Kpaa - 137.9 Kpaa)
Now, we convert the pressure values from Kpaa to KPa (1 Kpaa = 0.09807 KPa):
ΔT = (689.5 - 137.9) * 0.09807 KPa
ΔU = 1.36 kg * 0.7186 kJ/(kg·K) * ΔT
Calculate the value of ΔU.
(b) Δ(pV):
Using the formula Δ(pV) = p2V2 - p1V1, we can find the change in pressure and volume.
Given:
p1 = 137.9 Kpaa * 0.09807 KPa/Kpaa
p2 = 689.5 Kpaa * 0.09807 KPa/Kpaa
V1 = 0.142 m^3
V2 = 0.142 m^3
Calculate the value of Δ(pV).
(c) Heat Transfer (Q):
Using the equation Q = ΔU + W, we can calculate the heat transfer.
Given that Q = 0 (since the problem states the system is insulated) and ΔU = -W (from part (a)), we find:
Q = ΔU + W
0 = -W + W
0 = 0
Therefore, Q = 0.
To summarize:
(a) The work input (W) can be determined using the formulas mentioned above.
(b) The change in pressure and volume Δ(pV) can be calculated using the provided values.
(c) The heat transfer (Q) is zero since the system is insulated.