To solve the problem step by step, we will follow the outlined process to calculate the final temperature of the water after the gas expansion.
### Step 1: Calculate the Work Done by the Gas
The work done (W) during the expansion of a gas at constant pressure can be calculated using the formula:
\[ W = P \Delta V \]
Where:
- \( P \) = pressure in atm
- \( \Delta V \) = change in volume in dm³
Given:
- Initial volume \( V_1 = 3 \, \text{dm}^3 \)
- Final volume \( V_2 = 5 \, \text{dm}^3 \)
- Pressure \( P = 3 \, \text{atm} \)
First, we calculate \( \Delta V \):
\[ \Delta V = V_2 - V_1 = 5 \, \text{dm}^3 - 3 \, \text{dm}^3 = 2 \, \text{dm}^3 \]
Now, substituting the values into the work formula:
\[ W = 3 \, \text{atm} \times 2 \, \text{dm}^3 = 6 \, \text{atm} \cdot \text{dm}^3 \]
To convert this to Joules, we use the conversion factor \( 1 \, \text{atm} \cdot \text{dm}^3 = 101.3 \, \text{J} \):
\[ W = 6 \, \text{atm} \cdot \text{dm}^3 \times 101.3 \, \text{J/(atm} \cdot \text{dm}^3) = 607.8 \, \text{J} \]
### Step 2: Relate Work Done to Heat Transfer
According to the first law of thermodynamics, the work done by the gas is equal to the heat absorbed by the water:
\[ Q = W \]
So, \( Q = 607.8 \, \text{J} \).
### Step 3: Calculate the Change in Temperature of Water
The heat absorbed by the water can also be expressed using the specific heat capacity formula:
\[ Q = m \cdot C \cdot \Delta T \]
Where:
- \( m \) = mass of water in grams
- \( C \) = specific heat capacity of water
- \( \Delta T \) = change in temperature in Kelvin
Given:
- Number of moles of water \( n = 10 \, \text{mol} \)
- Molar mass of water \( = 18 \, \text{g/mol} \)
- Specific heat of water \( C = 4.184 \, \text{J/(g} \cdot \text{K)} \)
First, calculate the mass of water:
\[ m = n \times \text{molar mass} = 10 \, \text{mol} \times 18 \, \text{g/mol} = 180 \, \text{g} \]
Now, substituting the values into the heat equation:
\[ 607.8 \, \text{J} = 180 \, \text{g} \times 4.184 \, \text{J/(g} \cdot \text{K)} \times \Delta T \]
### Step 4: Solve for \( \Delta T \)
Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{607.8 \, \text{J}}{180 \, \text{g} \times 4.184 \, \text{J/(g} \cdot \text{K)}} \]
Calculating the denominator:
\[ 180 \times 4.184 = 752.32 \, \text{J/K} \]
Now substituting back:
\[ \Delta T = \frac{607.8}{752.32} \approx 0.807 \, \text{K} \]
### Step 5: Calculate Final Temperature of Water
The initial temperature of the water is given as \( T_1 = 290 \, \text{K} \). Therefore, the final temperature \( T_2 \) can be calculated as:
\[ T_2 = T_1 + \Delta T = 290 \, \text{K} + 0.807 \, \text{K} \approx 290.807 \, \text{K} \]
### Final Answer
The final temperature of the water is approximately:
\[ T_2 \approx 290.81 \, \text{K} \]
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