A graph is plotted for a vanderwaal's gas between `PV_(m)` vs P leading to an intercept of 22.16 litre-atm. The temperature of gas at which these observations of P and `V_(m)` were made is ___________ `""^(@)C`
`(R = 0.08 " litre atm" K^(-1) mol^(-1))`

Explanation

To solve the problem, we need to use the Van der Waals equation and the information given about the graph of \( PV_m \) vs \( P \).

### Step-by-Step Solution:

1. **Understand the Van der Waals Equation:**
The Van der Waals equation for one mole of gas is given by:
\[ \left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT \]
where \( P \) is the pressure, \( V_m \) is the molar volume, \( a \) and \( b \) are Van der Waals constants, \( R \) is the universal gas constant, and \( T \) is the temperature.

2. **Rearranging the Equation:**
For high pressures, we can rearrange the equation to express \( PV_m \):
\[ PV_m = RT - \frac{a}{V_m} \]
In this case, we can assume \( n = 1 \) (one mole of gas).

3. **Graph Interpretation:**
The graph plotted is \( PV_m \) vs \( P \). This can be rearranged to the form of a linear equation \( y = mx + c \):
\[ PV_m = -\frac{a}{V_m} + RT \]
Here, \( y = PV_m \), \( x = P \), \( m = -\frac{1}{V_m} \), and \( c = RT \).

4. **Using the Intercept:**
The intercept \( c \) of the graph is given as 22.16 L-atm. This represents \( RT \):
\[ RT = 22.16 \, \text{L-atm} \]

5. **Substituting the Value of R:**
We are given \( R = 0.08 \, \text{L-atm K}^{-1} \text{mol}^{-1} \). To find the temperature \( T \), we can rearrange the equation:
\[ T = \frac{RT}{R} = \frac{22.16}{0.08} \]

6. **Calculating Temperature:**
Now, perform the calculation:
\[ T = \frac{22.16}{0.08} = 277 \, \text{K} \]

7. **Converting Kelvin to Celsius:**
To convert the temperature from Kelvin to Celsius, use the formula:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T(°C) = 277 - 273 = 4 \, °C \]

### Final Answer:
The temperature of the gas at which these observations of \( P \) and \( V_m \) were made is **4 °C**.