One gram mole of a monoatomic gas occupies 22.4 L at S.T.P

Answer

To solve the problem of proving that 1 gram mole of a monoatomic gas occupies 22.4 L at standard temperature and pressure (STP), we can use the ideal gas law equation. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Identify the Conditions at STP
At standard temperature and pressure (STP):
- Temperature \( T = 273 \, \text{K} \)
- Pressure \( P = 1 \, \text{atm} \)

### Step 3: Substitute Values into the Ideal Gas Law
Given that we have 1 gram mole of a monoatomic gas, we have:
- \( n = 1 \, \text{mol} \)

The ideal gas constant \( R \) is:
\[ R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 4: Rearrange the Ideal Gas Law to Solve for Volume
We need to find the volume \( V \) at STP:
\[ V = \frac{nRT}{P} \]

### Step 5: Substitute the Known Values
Now, substitute the known values into the equation:
\[ V = \frac{(1 \, \text{mol}) \times (0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (273 \, \text{K})}{1 \, \text{atm}} \]

### Step 6: Calculate the Volume
Now perform the calculation:
\[ V = \frac{(1) \times (0.082) \times (273)}{1} \]
\[ V = 0.082 \times 273 \]
\[ V = 22.416 \, \text{L} \]

### Step 7: Conclusion
Thus, we can conclude that 1 gram mole of a monoatomic gas occupies approximately 22.4 L at STP.