A rigid tank of small mass contains 40.0 g of argon, initially at 200°C and 100 kPa. The tank is placed into a reservoir at 0°C and allowed to cool to thermal equilibrium. (a) Calculate the volume of the tank. (b) Calculate the change in internal energy of the argon. (c) Calculate the energy transferred by heat. (d) Calculate the change in entropy of the argon. (e) Calculate the change in entropy of the constant- temperature bath.
Well, this question seems to have a lot of hot and cool stuff going on. Let's see if I can make sense of it for you!
(a) To calculate the volume of the tank, you need to use the ideal gas law equation. But let's not make it too complicated, otherwise, we might have a meltdown! So, you can use the formula V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume (which we're trying to find), and T2 is the final temperature.
(b) The change in internal energy of the argon can be calculated using the equation ΔU = nCvΔT. But don't let the fancy symbols scare you! ΔU represents the change in internal energy, n is the number of moles (which you can find using the given mass and molar mass of argon), Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature.
(c) To calculate the energy transferred by heat, you need to use the equation Q = nCpΔT. Q here represents the heat transferred, n is again the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is, once again, the change in temperature.
(d) Calculating the change in entropy of the argon requires using the equation ΔS = nCp ln(T2/T1). Here, ΔS represents the change in entropy, n is the number of moles, Cp is the molar heat capacity at constant pressure, and ln represents the natural logarithm.
(e) Lastly, to calculate the change in entropy of the constant-temperature bath, you can use the equation ΔS(bath) = -ΔS.
I hope these calculations don't give you a brain freeze! But if you have any other questions, just let me know, and I'll try to keep it cool!
To solve this problem, we can use the ideal gas law and thermodynamic principles.
(a) To calculate the volume of the tank, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin:
T1 = 200°C + 273.15 = 473.15 K
Given:
P1 = 100 kPa = 100,000 Pa
n = 40.0 g of argon
R = 8.314 J/(mol·K)
Using the ideal gas law equation, we can solve for V:
V1 = (n * R * T1) / P1
V1 = (0.04 kg / 39.948 kg/mol) * 8.314 J/(mol·K) * 473.15 K / 100,000 Pa
V1 ≈ 0.0198 m^3
Therefore, the initial volume of the tank is approximately 0.0198 m^3.
(b) The change in internal energy in this case is equal to the heat transferred, as there is no work done. The change in internal energy can be calculated using the equation:
ΔU = Q
Where ΔU is the change in internal energy and Q is the heat transferred.
(c) The energy transferred by heat can be calculated using the formula:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m = 40.0 g = 0.04 kg (mass of argon)
c = specific heat capacity for argon = 0.52 kJ/(kg·K)
ΔT = 200°C - 0°C = 200 K
Converting the specific heat capacity to SI units:
c = 0.52 kJ/(kg·K) * 1000 J/(kJ) = 520 J/(kg·K)
Calculating the heat transferred:
Q = (0.04 kg) * (520 J/(kg·K)) * (200 K) = 4160 J
Therefore, the energy transferred by heat is 4160 J.
(d) The change in entropy of the argon can be calculated using the equation:
ΔS = mc ln(T2/T1)
Where ΔS is the change in entropy, m is the mass, c is the specific heat capacity, T2 is the final temperature, and T1 is the initial temperature.
Given:
T2 = 0°C + 273.15 = 273.15 K
Calculating the change in entropy:
ΔS = (0.04 kg) * (520 J/(kg·K)) * ln(273.15 K / 473.15 K)
(e) The change in entropy of the constant-temperature bath is given by:
ΔS_bath = -ΔS_system
Therefore, the change in entropy of the constant-temperature bath is equal to the negative change in entropy of the argon.
Note: To calculate the change in entropy, we need to know whether the process is reversible. We assumed a reversible process for simplicity in this explanation.
To answer the given questions, we need to use the ideal gas law, the specific heat capacity formulas, and the definition of entropy change.
(a) To calculate the volume of the tank, we can use the ideal gas law, which states that:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the given temperature from Celsius to Kelvin:
T_initial = 200°C + 273.15 = 473.15 K
Using the given pressure (100 kPa) and the molar mass of argon (40.0 g/mol), we can calculate the number of moles of argon:
n = mass / molar mass
n = 40.0 g / 40.0 g/mol
n = 1 mol
Now we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
V = (1 mol * (8.314 J/(mol·K)) * 473.15 K) / (100 kPa * 1000 Pa/kPa)
V = 19.8 L
Therefore, the volume of the tank is 19.8 liters.
(b) The change in internal energy of the argon can be calculated using the equation:
ΔU = ncΔT
where ΔU is the change in internal energy, n is the number of moles of gas, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity of argon at constant volume is approximately 0.520 J/(g·K).
ΔT = T_final - T_initial
ΔT = (0°C + 273.15 K) - (200°C + 273.15 K)
ΔT = 70.00 K
ΔU = (1 mol * 0.520 J/(g·K) * 70.00 K)
ΔU = 36.40 J
The change in internal energy of the argon is 36.40 Joules.
(c) The energy transferred by heat can be calculated using the equation:
Q = ΔU
Since the change in internal energy (ΔU) is the energy transferred by heat at constant volume, the energy transferred by heat is 36.40 Joules.
(d) To calculate the change in entropy of the argon, we can use the equation:
ΔS = nc ln(T_final / T_initial)
where ΔS is the change in entropy, n is the number of moles of gas, c is the molar heat capacity at constant volume, and ln represents the natural logarithm.
The molar heat capacity of argon at constant volume is approximately 12.47 J/(mol·K).
ΔS = (1 mol * 12.47 J/(mol·K) * ln((0°C + 273.15 K) / (200°C + 273.15 K)))
ΔS = -7.64 J/K
Therefore, the change in entropy of the argon is -7.64 Joules per Kelvin.
(e) The change in entropy of the constant-temperature bath is equal to the amount of heat transferred divided by the temperature of the bath:
ΔS_bath = Q / T_bath
The temperature of the constant-temperature bath is 0°C + 273.15 K = 273.15 K.
ΔS_bath = 36.40 J / 273.15 K
ΔS_bath = 0.133 J/K
Therefore, the change in entropy of the constant-temperature bath is 0.133 Joules per Kelvin.