A sample of oxygen gas expands its volume from `3L` to `5L` against a constant pressure of `3` atm. If work done during expansion be used to heat `10 ` mole of water initially present at `290K` , its finally temperature will be `(` specific heat capacity of water `=4.18 J//k-g) :`

Options

A. 292.0 K
B. 298.0 K
C. 290.8 K
D. 293.7 K

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the Work Done (W)
The work done during the expansion of gas at constant pressure can be calculated using the formula:
\[ W = -P_{\text{external}} \times (V_2 - V_1) \]

Where:
- \( P_{\text{external}} = 3 \, \text{atm} \)
- \( V_1 = 3 \, \text{L} \)
- \( V_2 = 5 \, \text{L} \)

Substituting the values:
\[ W = -3 \, \text{atm} \times (5 \, \text{L} - 3 \, \text{L}) \]
\[ W = -3 \, \text{atm} \times 2 \, \text{L} \]
\[ W = -6 \, \text{atm} \cdot \text{L} \]

### Step 2: Convert Work Done to Joules
To convert the work done from atm·L to Joules, we use the conversion factor:
\[ 1 \, \text{atm} \cdot \text{L} = 101.3 \, \text{J} \]

Thus,
\[ W = -6 \, \text{atm} \cdot \text{L} \times 101.3 \, \text{J/atm·L} \]
\[ W = -607.8 \, \text{J} \]

### Step 3: Relate Work Done to Heat Supplied (Q)
The work done on the system is equal to the heat supplied to the water, which can be expressed as:
\[ Q = mc\Delta T \]

Where:
- \( m \) = mass of water
- \( c \) = specific heat capacity of water = \( 4.18 \, \text{J/g·K} \)
- \( \Delta T \) = change in temperature

### Step 4: Calculate the Mass of Water
Given that we have 10 moles of water, we can calculate the mass:
- Molar mass of water = 18 g/mol
\[ m = 10 \, \text{moles} \times 18 \, \text{g/mol} = 180 \, \text{g} \]

### Step 5: Substitute Values into the Heat Equation
Now substituting the values into the heat equation:
\[ -607.8 \, \text{J} = 180 \, \text{g} \times 4.18 \, \text{J/g·K} \times \Delta T \]

### Step 6: Solve for ΔT
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{-607.8 \, \text{J}}{180 \, \text{g} \times 4.18 \, \text{J/g·K}} \]
\[ \Delta T = \frac{-607.8}{752.4} \]
\[ \Delta T \approx -0.81 \, \text{K} \]

### Step 7: Calculate Final Temperature
The change in temperature \( \Delta T \) is related to the initial temperature \( T_i \):
\[ T_f = T_i + \Delta T \]
Where \( T_i = 290 \, \text{K} \):
\[ T_f = 290 \, \text{K} - 0.81 \, \text{K} \]
\[ T_f \approx 289.19 \, \text{K} \]

### Final Answer
The final temperature of the water after the expansion of the gas is approximately:
\[ T_f \approx 289.19 \, \text{K} \]

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