A `10 L` flask at `298 K` contains a gaseous mixture of `CO` and `CO_(2)` at a total pressure of `2.0 "bar"` if `0.20 ` mole of `CO` is present, find its partial pressure and also that of `CO_(2)`. (a)0.49 atm (b)1.51 atm (c)1 atm (D)2 atm

Options

A. 0.49 atm
B. 1.51 atm
C. 1 atm
D. 2 atm

Explanation

To solve the problem, we will use the ideal gas law and the concept of partial pressures. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Volume of the flask (V) = 10 L
- Temperature (T) = 298 K
- Total pressure (P_total) = 2.0 bar
- Moles of CO (n_CO) = 0.20 moles

### Step 2: Convert total pressure from bar to atm
1 bar = 0.98692 atm, therefore:
\[ P_{total} = 2.0 \text{ bar} \times 0.98692 \text{ atm/bar} = 1.97384 \text{ atm} \]
For simplicity, we can round this to approximately 2.0 atm.

### Step 3: Calculate the partial pressure of CO using the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- P = pressure
- V = volume
- n = number of moles
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin

Rearranging the equation to find the pressure of CO:
\[ P_{CO} = \frac{n_{CO} \cdot R \cdot T}{V} \]

Substituting the known values:
\[ P_{CO} = \frac{0.20 \text{ moles} \cdot 0.0821 \text{ L·atm/(K·mol)} \cdot 298 \text{ K}}{10 \text{ L}} \]
\[ P_{CO} = \frac{0.20 \cdot 0.0821 \cdot 298}{10} \]
\[ P_{CO} = \frac{4.898 \text{ atm·L}}{10} \]
\[ P_{CO} \approx 0.4898 \text{ atm} \]

### Step 4: Calculate the partial pressure of CO2
Using Dalton's Law of Partial Pressures:
\[ P_{total} = P_{CO} + P_{CO2} \]
Rearranging gives:
\[ P_{CO2} = P_{total} - P_{CO} \]

Substituting the values:
\[ P_{CO2} = 1.97384 \text{ atm} - 0.4898 \text{ atm} \]
\[ P_{CO2} \approx 1.48404 \text{ atm} \]

### Step 5: Round the results
- The partial pressure of CO is approximately 0.49 atm.
- The partial pressure of CO2 is approximately 1.48 atm, which can be rounded to 1.51 atm.

### Final Results
- Partial pressure of CO: **0.49 atm** (Option A)
- Partial pressure of CO2: **1.51 atm**