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 determine the temperature of a Van der Waals gas from the given intercept of the graph plotted between \( PV_m \) (pressure multiplied by molar volume) and \( P \) (pressure). The intercept is given as 22.16 liter-atm, and the gas constant \( R \) is 0.08 liter-atm K\(^{-1}\) mol\(^{-1}\).

### Step-by-Step Solution:

1. **Understand the Van der Waals Equation:**
The Van der Waals equation for one mole of gas can be written as:
\[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \]
where \( a \) and \( b \) are Van der Waals constants.

2. **Simplify for High Pressure:**
At high pressure, the term \( \frac{a}{V_m^2} \) becomes negligible, and we can simplify the equation to:
\[ P(V_m - b) = RT \]
Rearranging gives:
\[ PV_m - Pb = RT \]
or
\[ PV_m = Pb + RT \]

3. **Identify the Intercept:**
From the equation \( PV_m = Pb + RT \), we can see that if we plot \( PV_m \) against \( P \), the y-intercept (when \( P = 0 \)) is equal to \( RT \).

4. **Set Up the Equation:**
Given that the intercept is 22.16 liter-atm, we can write:
\[ RT = 22.16 \]

5. **Substitute the Value of R:**
We know \( R = 0.08 \) liter-atm K\(^{-1}\) mol\(^{-1}\). Substituting this into the equation gives:
\[ 0.08T = 22.16 \]

6. **Solve for Temperature (T):**
To find \( T \), we rearrange the equation:
\[ T = \frac{22.16}{0.08} \]
Calculating this gives:
\[ T = 277 \text{ K} \]

7. **Convert Temperature to Celsius:**
To convert from Kelvin to Celsius, we 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 were made is **4 °C**.