Answer
The energy that becomes unavailable to do work when hydrogen leaks out of the fuel tank into a vacuum can be calculated using the following equation:
ΔU = q - w
where ΔU is the change in internal energy of the gas, q is the heat added to or removed from the gas, and w is the work done by or on the gas.
In this case, the gas is expanding into a vacuum, so the work done by the gas is equal to the change in its internal energy. Therefore, we can write:
ΔU = -w
The energy that becomes unavailable to do work is equal to the change in internal energy of the gas, so we can set ΔU equal to the given value of 54.6 MJ:
ΔU = 54.6 MJ = -w
We can calculate the initial number of moles of hydrogen in the fuel tank using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Rearranging this equation, we get:
n = PV / RT
Substituting the given values, we get:
n = (70 MPa)(5.0 kg) / (8.31 J/mol K)(293 K)
n = 0.105 mol
This is the initial number of moles of hydrogen in the fuel tank. We can assume that this number of moles remains constant as the gas leaks out into the vacuum chamber.
We can calculate the final volume of the gas by using the ideal gas law again:
PV = nRT
Rearranging this equation, we get:
V = nRT / P
Substituting the given values, we get:
V = (0.105 mol)(8.31 J/mol K)(293 K) / (70 MPa)
V = 4.34 m3
This is the final volume of the gas in the vacuum chamber. To find the volume of the fuel tank, we need to subtract the initial volume of the gas from the final volume:
fuel tank volume = final volume - initial volume
fuel tank volume = 4.34 m3 - V_H2
where V_H2 is the volume occupied by 5.0 kg of hydrogen at 70 MPa and 293 K. We can calculate this volume using the ideal gas law:
PV = nRT
Rearranging this equation, we get:
V = nRT / P
Substituting the given values and solving for V, we get:
V_H2 = (5.0 kg / 2.016 g/mol)(8.31 J/mol K)(293 K) / (70 MPa)
V_H2 = 0.174 m3
Substituting this value into the equation for the fuel tank volume, we get:
fuel tank volume = 4.34 m3 - 0.174 m3
fuel tank volume = 4.16 m3
Therefore, the volume of the fuel tank is approximately 4.16 m3.
Work Step by Step
The energy that becomes unavailable to do work when hydrogen leaks out of the fuel tank into a vacuum can be calculated using the following equation:
ΔU = q - w
where ΔU is the change in internal energy of the gas, q is the heat added to or removed from the gas, and w is the work done by or on the gas.
In this case, the gas is expanding into a vacuum, so the work done by the gas is equal to the change in its internal energy. Therefore, we can write:
ΔU = -w
The energy that becomes unavailable to do work is equal to the change in internal energy of the gas, so we can set ΔU equal to the given value of 54.6 MJ:
ΔU = 54.6 MJ = -w
We can calculate the initial number of moles of hydrogen in the fuel tank using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Rearranging this equation, we get:
n = PV / RT
Substituting the given values, we get:
n = (70 MPa)(5.0 kg) / (8.31 J/mol K)(293 K)
n = 0.105 mol
This is the initial number of moles of hydrogen in the fuel tank. We can assume that this number of moles remains constant as the gas leaks out into the vacuum chamber.
We can calculate the final volume of the gas by using the ideal gas law again:
PV = nRT
Rearranging this equation, we get:
V = nRT / P
Substituting the given values, we get:
V = (0.105 mol)(8.31 J/mol K)(293 K) / (70 MPa)
V = 4.34 m3
This is the final volume of the gas in the vacuum chamber. To find the volume of the fuel tank, we need to subtract the initial volume of the gas from the final volume:
fuel tank volume = final volume - initial volume
fuel tank volume = 4.34 m3 - V_H2
where V_H2 is the volume occupied by 5.0 kg of hydrogen at 70 MPa and 293 K. We can calculate this volume using the ideal gas law:
PV = nRT
Rearranging this equation, we get:
V = nRT / P
Substituting the given values and solving for V, we get:
V_H2 = (5.0 kg / 2.016 g/mol)(8.31 J/mol K)(293 K) / (70 MPa)
V_H2 = 0.174 m3
Substituting this value into the equation for the fuel tank volume, we get:
fuel tank volume = 4.34 m3 - 0.174 m3
fuel tank volume = 4.16 m3
Therefore, the volume of the fuel tank is approximately 4.16 m3.
