To determine the number of correct statements among the provided options, we will analyze each statement one by one.
### Step 1: Analyze Statement 1
**Statement:** The expansion work for a gas into vacuum is equal to zero.
**Solution:**
In thermodynamics, when a gas expands into a vacuum, there is no opposing pressure. The work done (W) is calculated using the formula:
\[ W = -P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume. Since the pressure \( P \) in a vacuum is zero, the work done is:
\[ W = -0 \times \Delta V = 0 \]
Thus, this statement is **correct**.
### Step 2: Analyze Statement 2
**Statement:** 1 mole of a gas occupying 3 liters volume on expanding to 15 liters at constant pressure of 1 atm does expansion work 1.215 kJ.
**Solution:**
To calculate the work done, we use:
\[ W = -P \Delta V \]
where:
- \( P = 1 \, \text{atm} \)
- \( \Delta V = V_2 - V_1 = 15 \, \text{L} - 3 \, \text{L} = 12 \, \text{L} \)
Now, converting the work into joules:
\[ W = -1 \, \text{atm} \times 12 \, \text{L} \]
Using the conversion \( 1 \, \text{L atm} = 101.325 \, \text{J} \):
\[ W = -12 \times 101.325 \, \text{J} = -1215.9 \, \text{J} = -1.215 \, \text{kJ} \]
Thus, this statement is also **correct**.
### Step 3: Analyze Statement 3
**Statement:** The maximum work done by the gas during reversible expansion of 16g O2 at 300K from 5dm³ to 25dm³ is 2.0074 kJ.
**Solution:**
To find the maximum work done, we use the formula for reversible work:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Where:
- \( n = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)
- \( V_1 = 5 \, \text{dm}^3 \) and \( V_2 = 25 \, \text{dm}^3 \)
Calculating:
\[ W = -0.5 \times 8.314 \times 300 \times \ln\left(\frac{25}{5}\right) \]
\[ W = -0.5 \times 8.314 \times 300 \times \ln(5) \]
Calculating \( \ln(5) \approx 1.609 \):
\[ W \approx -0.5 \times 8.314 \times 300 \times 1.609 \]
\[ W \approx -2.0074 \, \text{kJ} \]
The statement is incorrect because it does not mention the negative sign. Therefore, this statement is **incorrect**.
### Step 4: Analyze Statement 4
**Statement:** The ΔS for s → l is almost negligible in comparison to ΔS for l → g.
**Solution:**
The change in entropy (ΔS) for the transition from solid to liquid is indeed much smaller than the change in entropy for the transition from liquid to gas. This is because solids have more ordered structures compared to liquids, while gases have a much higher degree of disorder. Thus, this statement is **correct**.
### Step 5: Analyze Statement 5
**Statement:** ΔS = 2.303 nR log(V2/V1) (at constant T).
**Solution:**
This statement is correct as it represents the change in entropy for an ideal gas during an isothermal expansion or compression. Thus, this statement is **correct**.
### Step 6: Analyze Statement 6
**Statement:** Reversible isothermal work done = -2.303nRT log10(V2/V1).
**Solution:**
This statement is also correct as it reflects the formula for work done during a reversible isothermal expansion. Thus, this statement is **correct**.
### Conclusion
After analyzing all statements, we find that statements 1, 2, 4, 5, and 6 are correct, while statement 3 is incorrect. Therefore, the number of correct statements is **5**.
