If 2 moles each of CO, `N_(2)` and `CO_(2)` were taken in a 5 litre vessel at 300K and the entire `CO_(2)` was absorbed into KOH, then the partial pressure exerted by CO in the vessel after absorption will be

Options

A. 9.852 atm
B. 29.55 atm
C. 19.22 atm
D. 40 atm

Answer

To solve the problem step by step, we will follow the outlined approach to find the partial pressure of carbon monoxide (CO) after the absorption of carbon dioxide (CO₂) by KOH.

### Step 1: Identify the initial conditions
We have:
- 2 moles of CO
- 2 moles of N₂
- 2 moles of CO₂
- Volume of the vessel = 5 L
- Temperature = 300 K

### Step 2: Calculate the total initial moles
Initially, we have:
- Total moles = moles of CO + moles of N₂ + moles of CO₂
- Total moles = 2 + 2 + 2 = 6 moles

### Step 3: Determine the moles after CO₂ absorption
Since all the CO₂ is absorbed by KOH, the remaining moles in the vessel will be:
- Remaining moles = moles of CO + moles of N₂
- Remaining moles = 2 + 2 = 4 moles

### Step 4: Calculate the total pressure using the ideal gas law
The ideal gas law is given by the formula:
\[ PV = nRT \]
Where:
- P = pressure (in atm)
- V = volume (in L)
- n = number of moles
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature (in K)

Rearranging for pressure (P):
\[ P = \frac{nRT}{V} \]

Substituting the values:
- n = 4 moles (after CO₂ absorption)
- R = 0.0821 L·atm/(K·mol)
- T = 300 K
- V = 5 L

Calculating the total pressure:
\[ P = \frac{4 \times 0.0821 \times 300}{5} \]
\[ P = \frac{98.52}{5} \]
\[ P = 19.704 \, \text{atm} \]

### Step 5: Calculate the mole fraction of CO
The mole fraction (X) of CO is given by:
\[ X_{CO} = \frac{\text{moles of CO}}{\text{total moles}} \]
\[ X_{CO} = \frac{2}{4} = 0.5 \]

### Step 6: Calculate the partial pressure of CO
The partial pressure (P_CO) of CO can be calculated using:
\[ P_{CO} = X_{CO} \times P_{total} \]
Substituting the values:
\[ P_{CO} = 0.5 \times 19.704 \]
\[ P_{CO} = 9.852 \, \text{atm} \]

### Final Answer
The partial pressure exerted by CO in the vessel after absorption is **9.852 atm**.

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