A litre of dry air weighs 1.293 gram at S.T.P Find the temperature at which a litre of air will weigh one gram when the pressure is 72 cm. of mercury.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between mass, volume, temperature, and pressure. The ideal gas equation can be expressed as:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 1: Identify the known values at STP
At Standard Temperature and Pressure (STP):
- Mass of 1 litre of air, \( m_1 = 1.293 \, \text{g} \)
- Temperature, \( T_1 = 273 \, \text{K} \)
- Pressure, \( P_1 = 76 \, \text{cm Hg} \) (standard atmospheric pressure)

### Step 2: Identify the conditions for the second scenario
In the second scenario:
- Mass of 1 litre of air, \( m_2 = 1 \, \text{g} \)
- Volume, \( V_2 = 1 \, \text{litre} \)
- Pressure, \( P_2 = 72 \, \text{cm Hg} \)
- We need to find the temperature \( T_2 \).

### Step 3: Use the relationship between the two conditions
Since the molecular weight of air remains constant, we can use the following relationship derived from the ideal gas law:

\[ \frac{m_1 \cdot T_1}{P_1} = \frac{m_2 \cdot T_2}{P_2} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:

\[ \frac{1.293 \, \text{g} \cdot 273 \, \text{K}}{76 \, \text{cm Hg}} = \frac{1 \, \text{g} \cdot T_2}{72 \, \text{cm Hg}} \]

### Step 5: Rearranging the equation to solve for \( T_2 \)
Rearranging gives:

\[ T_2 = \frac{1.293 \cdot 273 \cdot 72}{76} \]

### Step 6: Calculate \( T_2 \)
Calculating the right side:

\[ T_2 = \frac{1.293 \cdot 273 \cdot 72}{76} \approx 334.4 \, \text{K} \]

### Step 7: Convert \( T_2 \) to Celsius
To convert from Kelvin to Celsius:

\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 = 334.4 - 273 \approx 61.4 \, \text{°C} \]

### Final Answer
The temperature at which a litre of air will weigh one gram when the pressure is 72 cm of mercury is approximately **61.4 °C**.

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