To find the work done by the gas during an adiabatic expansion, we can use the formula:

W = (γ / (γ - 1)) * P_initial * (V_final - V_initial)

Where:
W is the work done by the gas
γ is the heat capacity ratio (for a monoatomic gas, γ = 5/3)
P_initial is the initial pressure of the gas
V_final is the final volume of the gas
V_initial is the initial volume of the gas

Given that the gas is ideal and monoatomic, we know that the pressure and volume are related by the equation:

P * V = n * R * T

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas

We are given that the gas occupies a volume of V at a temperature of 27 degrees Celsius. To find the initial pressure, we need to convert the temperature to Kelvin:

T_initial = 27 + 273 = 300 K

Since we have two moles of gas, we can substitute these values into the ideal gas equation to find the initial pressure:

P_initial * V = 2 * R * T_initial

Now, we are told that the gas is expanded adiabatically to a volume of 2V. This means that the final volume is 2V, and the final pressure can be found using the ideal gas equation:

P_final * (2V) = 2 * R * T_initial

Now we have all the necessary values to calculate the work done by the gas:

W = (γ / (γ - 1)) * P_initial * (V_final - V_initial)
W = (5/3 / (5/3 - 1)) * P_initial * (2V - V)
W = (5/3 / (2/3)) * P_initial * V
W = (5/2) * P_initial * V

Substituting the value of P_initial * V from the ideal gas equation:

W = (5/2) * (2 * R * T_initial)
W = 5 * R * T_initial

Finally, we can substitute the value of T_initial:

W = 5 * R * 300

Using the value of the ideal gas constant R = 8.314 J/(mol·K), we can calculate the work done by the gas:

W = 5 * 8.314 * 300 = 12471 J

Therefore, the work done by the gas in the SI system is 12471 Joules.

Question

To find the work done by the gas during an adiabatic expansion, we can use the formula:

W = (γ / (γ - 1)) * P_initial * (V_final - V_initial)

Where:
W is the work done by the gas
γ is the heat capacity ratio (for a monoatomic gas, γ = 5/3)
P_initial is the initial pressure of the gas
V_final is the final volume of the gas
V_initial is the initial volume of the gas

Given that the gas is ideal and monoatomic, we know that the pressure and volume are related by the equation:

P * V = n * R * T

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas

We are given that the gas occupies a volume of V at a temperature of 27 degrees Celsius. To find the initial pressure, we need to convert the temperature to Kelvin:

T_initial = 27 + 273 = 300 K

Since we have two moles of gas, we can substitute these values into the ideal gas equation to find the initial pressure:

P_initial * V = 2 * R * T_initial

Now, we are told that the gas is expanded adiabatically to a volume of 2V. This means that the final volume is 2V, and the final pressure can be found using the ideal gas equation:

P_final * (2V) = 2 * R * T_initial

Now we have all the necessary values to calculate the work done by the gas:

W = (γ / (γ - 1)) * P_initial * (V_final - V_initial)
W = (5/3 / (5/3 - 1)) * P_initial * (2V - V)
W = (5/3 / (2/3)) * P_initial * V
W = (5/2) * P_initial * V

Substituting the value of P_initial * V from the ideal gas equation:

W = (5/2) * (2 * R * T_initial)
W = 5 * R * T_initial

Finally, we can substitute the value of T_initial:

W = 5 * R * 300

Using the value of the ideal gas constant R = 8.314 J/(mol·K), we can calculate the work done by the gas:

W = 5 * 8.314 * 300 = 12471 J

Therefore, the work done by the gas in the SI system is 12471 Joules.

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