The density of air 380K and 722 mm of Hg is lg/`cm^3`. If air is cooled to 100K and 1 atm the final density is :

Answer

To solve the problem of finding the final density of air when it is cooled from 380 K and 722 mm of Hg to 100 K and 1 atm, we can use the ideal gas law and the relationship between density, pressure, and temperature. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert Initial Pressure to Atmospheres
The initial pressure \( P_1 \) is given as 722 mm of Hg. We need to convert this to atmospheres (atm) using the conversion factor \( 1 \text{ atm} = 760 \text{ mm of Hg} \).

\[ P_1 = \frac{722 \text{ mm of Hg}}{760 \text{ mm of Hg/atm}} = 0.95 \text{ atm} \]

**Hint:** Remember that to convert mm of Hg to atm, you divide by 760.

### Step 2: Identify Given Values
We have the following values:
- Initial Temperature \( T_1 = 380 \text{ K} \)
- Initial Density \( d_1 = 1 \text{ g/cm}^3 \)
- Final Temperature \( T_2 = 100 \text{ K} \)
- Final Pressure \( P_2 = 1 \text{ atm} \)

### Step 3: Use the Ideal Gas Law Relationship
According to the ideal gas law, the relationship between pressure, density, and temperature can be expressed as:

\[ \frac{P_1}{d_1 T_1} = \frac{P_2}{d_2 T_2} \]

Where:
- \( d_2 \) is the final density we need to find.

### Step 4: Rearrange the Equation to Solve for \( d_2 \)
Rearranging the equation gives us:

\[ d_2 = \frac{P_2 \cdot d_1 \cdot T_2}{P_1 \cdot T_1} \]

### Step 5: Substitute the Known Values
Now, we can substitute the known values into the equation:

\[ d_2 = \frac{(1 \text{ atm}) \cdot (1 \text{ g/cm}^3) \cdot (100 \text{ K})}{(0.95 \text{ atm}) \cdot (380 \text{ K})} \]

### Step 6: Calculate \( d_2 \)
Calculating the numerator and denominator:

\[ d_2 = \frac{100}{0.95 \cdot 380} \]

Calculating the denominator:

\[ 0.95 \cdot 380 = 361 \]

Now substituting back:

\[ d_2 = \frac{100}{361} \approx 0.277 \text{ g/cm}^3 \]

### Step 7: Final Calculation
Now, we can simplify this to find the final density:

\[ d_2 \approx 0.277 \text{ g/cm}^3 \]

### Conclusion
The final density of air when cooled to 100 K and 1 atm is approximately \( 0.277 \text{ g/cm}^3 \).

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