To analyze the adiabatic expansion of the gas, we need to apply the relevant thermodynamic principles.
- Work Done by the Gas (W):
For an adiabatic process, the work done by the gas can be calculated using the first law of thermodynamics, which states:
\[ \Delta U = Q - W \]
Where:
- \(\Delta U\) is the change in internal energy.
- \(Q\) is the heat added to the system (which is 0 for an adiabatic process).
- \(W\) is the work done by the system.
Since the process is adiabatic, \(Q = 0\), so the equation simplifies to:
\[ \Delta U = -W \]
To find the change in internal energy, we use the formula:
\[ \Delta U = nC_v \Delta T \]
Where:
- \(n\) is the number of moles (0.600 mol).
- \(C_v\) is the molar heat capacity at constant volume (for diatomic gases like O2, \(C_v \approx 5/2 R\) where \(R = 8.314 , \text{J/(mol·K)}\)).
- \(\Delta T\) is the change in temperature.
First, let's convert the temperatures from Celsius to Kelvin:
- \(T_i = 30°C = 303.15 , K\)
- \(T_f = 10°C = 283.15 , K\)
Now, calculate \(\Delta T\):
\[ \Delta T = T_f - T_i = 283.15 , K - 303.15 , K = -20 , K \]
Now, we calculate \(\Delta U\):
\[ C_v = \frac{5}{2} R = \frac{5}{2} \times 8.314 , \text{J/(mol·K)} \approx 20.786 , \text{J/(mol·K)} \]
\[ \Delta U = nC_v \Delta T = 0.600 , \text{mol} \times 20.786 , \text{J/(mol·K)} \times (-20 , K) \]
Calculating \(\Delta U\):
\[ \Delta U = 0.600 \times 20.786 \times (-20) \approx -249.432, \text{J} \]
Since \(\Delta U = -W\):
\[ W = -\Delta U = 249.432, \text{J} \]
Thus, the work done by the gas is approximately:
\[ W \approx 249.43 , \text{J} \]
- Heat Added to the Gas (Q):
As already established, for an adiabatic process, no heat is added or removed from the system:
\[ Q = 0 , \text{J} \]
Summary:
- The work done by the gas is approximately 249.43 J.
- The heat added to the gas is 0 J (since the process is adiabatic).