A ideal gas does work on its surroundings when it expands by 2.5 L against external pressure 2 atm. This work done is used to heat up 1 mole of water at 293 K. What would be the final temperature of water in kelvin if specific heat for water is `"4.184 Jg"^(-1)"K"^(-1)`?

Options

A. 300
B. 600
C. 200
D. 1000

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Work Done by the Gas
The work done (W) by the gas during expansion can be calculated using the formula:
\[ W = -P_{\text{external}} \times \Delta V \]
Where:
- \( P_{\text{external}} = 2 \, \text{atm} \)
- \( \Delta V = 2.5 \, \text{L} \)

Converting the work done from liters-atm to joules:
1 atm = 101.325 J, so:
\[ W = -2 \, \text{atm} \times 2.5 \, \text{L} = -5 \, \text{L-atm} \]
Now converting to joules:
\[ W = -5 \, \text{L-atm} \times 101.325 \, \text{J/L-atm} = -506.625 \, \text{J} \]
We can round this to:
\[ W \approx -506.3 \, \text{J} \]

### Step 2: Relate Work Done to Heat Supplied
The work done on the surroundings is equal to the heat supplied to the water:
\[ Q = -W = 506.3 \, \text{J} \]

### Step 3: Use the Heat Equation to Find Change in Temperature
The heat absorbed by the water can be expressed using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m = 18 \, \text{g} \) (mass of 1 mole of water)
- \( c = 4.184 \, \text{J/g} \cdot \text{K} \) (specific heat of water)
- \( \Delta T = T_{\text{final}} - T_{\text{initial}} \)

Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{Q}{m \cdot c} \]
Substituting the values:
\[ \Delta T = \frac{506.3 \, \text{J}}{18 \, \text{g} \times 4.184 \, \text{J/g} \cdot \text{K}} \]
Calculating:
\[ \Delta T = \frac{506.3}{75.312} \approx 6.72 \, \text{K} \]

### Step 4: Calculate the Final Temperature
Now, we can find the final temperature:
\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \]
Given that \( T_{\text{initial}} = 293 \, \text{K} \):
\[ T_{\text{final}} = 293 \, \text{K} + 6.72 \, \text{K} \approx 299.72 \, \text{K} \]

### Final Answer
Thus, the final temperature of the water is approximately:
\[ T_{\text{final}} \approx 300 \, \text{K} \]

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