High School

A 0.8-m³ rigid tank contains carbon dioxide (CO₂) gas at 250 K and 100 kPa. A 500-W electric resistance heater placed in the tank is now turned on and kept on for 40 minutes, after which the pressure of CO₂ is measured to be 175 kPa. Assuming the surroundings are at 300 K and using constant specific heats, determine:

(a) The final temperature of CO₂.

(b) The net amount of heat transfer from the tank.

(c) The entropy generation during this process.

Answer :

Answer:

Explanation:

Given:

Power, p = 500 W

Time, t = 40 min

= 2400 s

Volume, V = 0.8 m^3

Temperature, T = 250 K

Pressure, P = 100 kPa

Temperature of surroundings, Ts = 300 K

Using ideal gas equation,

PV = nRT

n = (100 × 10^3 × 0.8)/(8.3145 × 250)

= 38.49 mole

Mass = number of moles × molar mass

Molar mass = 12 + (16 × 2)

= 44 g/mol

Mass = 44 × 38.49

= 1693.43 g

= 1.693 kg.

A.

P2 = 175 kPa

Using pressure law,

P1/T1 = P2/T2

T2 = (175 × 250)/100

= 437.5 K

B.

Cvco2 = 0.706 kJ/kg.K

Total energy, U = Qin - Qout

Qout = (p × t) - (m × cv × delta T)

= (500 × 2400) - (1.693 × 0.706 × (437.5 - 250))

= 1200 kJ - 224.11 kJ

= 975.889 kJ

= 975.9 kJ

C.

Cpco2 = 0.895 kJ/kg.K

Gas constant, Rc = R/molar mass of CO2

= 8.3145/44

= 0.189

Using the formula ,

Entropy, S = (m × (Cpco2 × ln(T2/T1) - Rc × ln(P2/P1)) + Qout/Ts

Inputting values,

= (1.693 × (0.895 × ln(437.5/250) - 0.189 × ln(175/100)) + 975.9/300

= 3.922 kJ/K.

a) The final temperature of CO₂ is 450 K b) The net amount of heat transfer from the tank is 986.55 kJ c) The entropy generation during the process is -2288.59 J/K.

To solve this problem, we need to determine several key aspects:

(a) Final Temperature of CO₂

Given: V = 0.8 m³, P₁ = 100 kPa, T₁ = 250 K, P₂ = 175 kPa.

Using the ideal gas law equation, PV = nRT, we first determine the initial amount of moles of CO₂ (n):

  • P₁V = nRT₁,
    P1 = 100 kPa (100,000 Pa),
    R (specific gas constant for CO₂) = 8.314 J/mol·K.
    So, [tex]n = \frac{P_1 V}{RT_1} = \frac{100{,}000 \times 0.8}{8.314 \times 250} = 38.5 \text{ moles}[/tex]
  • This value remains constant. To find the final temperature (T₂):
    P₂V = nRT₂,
    So,
    [tex]T_2 = \frac{P_2 V}{nR} = \frac{175{,}000 \times 0.8}{38.5 \times 8.314} = 450 \, \text{K}[/tex]

(b) Net Amount of Heat Transfer

  • Given heater power: 500 W, time = 40 min (2400 seconds).
  • Energy added by the heater (Qin) = Power x time = 500 W x 2400 s = 1,200,000 J (1,200 kJ).
  • Using Q = mCΔT ,
  • where C (Cv, specific heat at constant volume) for CO₂ ≈ 28.46 J/mol·K
    [tex]Q = n C (T_2 - T_1)[/tex]
  • [tex]Q = 38.5 \times 28.46 \times (450 - 250) = 213{,}449 \, \text{J} \quad (213.45 \, \text{kJ})[/tex].
    Thus, net heat transfer is
  • Qnet = Qin - Q = 1,200 - 213.45 = 986.55 kJ

(c) Entropy Generation

  • [tex]\Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) = 38.5 \times 28.46 \times \ln\left(\frac{450}{250}\right) = 1711.41 \, \text{J/K}[/tex]
    With the surroundings at 300 K:
    [tex]S_{\text{surr}} = \frac{Q_{\text{in}}}{T_{\text{surroundings}}} = \frac{1{,}200{,}000}{300} = 4000 \, \text{J/K}[/tex]
  • [tex]S_{\text{gen}} = \Delta S_{\text{total}} - S_{\text{surr}} = 1711.41 - 4000 = -2288.59 \, \text{J/K}[/tex]

Therefore, a) The final temperature of CO₂ is 450 K b) The net amount of heat transfer from the tank is 986.55 kJ c) The entropy generation during the process is -2288.59 J/K.

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