For a certain gas `P_(c)` and `T_(c)` are `80atm` and `-33^(@)C` respectively then `V_(C)` of the gas will be approximately [Take `R=(1)/(2)" Ltr atm mol"^(-1)K^(-1)`]

Options

A. `(1)/(4)"Ltr mol"^(-1)`
B. `(1)/(16)"Ltr mol"^(-1)`
C. `(3)/(32)"Ltr mol"^(-1)`
D. None of these

Explanation

To solve the problem of finding the critical volume \( V_c \) of the gas given the critical pressure \( P_c \) and critical temperature \( T_c \), we can follow these steps:

### Step 1: Convert the Critical Temperature to Kelvin
The critical temperature is given as \( -33^\circ C \). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Substituting the value:
\[ T_c = -33 + 273.15 = 240.15 \, K \]

### Step 2: Use the Critical Condition Formula
From the critical conditions, we know that:
\[ \frac{P_c V_c}{R T_c} = \frac{3}{8} \]
Where:
- \( P_c = 80 \, atm \)
- \( R = \frac{1}{2} \, L \cdot atm \cdot mol^{-1} \cdot K^{-1} \)
- \( T_c = 240.15 \, K \)

### Step 3: Rearrange the Formula to Find \( V_c \)
Rearranging the equation to solve for \( V_c \):
\[ V_c = \frac{3}{8} \cdot \frac{R T_c}{P_c} \]

### Step 4: Substitute the Values into the Equation
Now we substitute the known values into the equation:
\[ V_c = \frac{3}{8} \cdot \frac{\left(\frac{1}{2} \, L \cdot atm \cdot mol^{-1} \cdot K^{-1}\right) \cdot (240.15 \, K)}{80 \, atm} \]

### Step 5: Calculate \( V_c \)
Calculating the numerator:
\[ \frac{1}{2} \cdot 240.15 = 120.075 \, L \cdot atm \cdot mol^{-1} \]
Now substituting this into the equation:
\[ V_c = \frac{3}{8} \cdot \frac{120.075}{80} \]
Calculating \( \frac{120.075}{80} \):
\[ \frac{120.075}{80} = 1.5009375 \]
Now substituting this back:
\[ V_c = \frac{3}{8} \cdot 1.5009375 = 0.5628515625 \]

### Step 6: Round to the Appropriate Significant Figures
Rounding this value gives:
\[ V_c \approx 0.56 \, L \]

### Final Answer
The critical volume \( V_c \) of the gas is approximately \( 0.56 \, L \).

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