By the ideal gas law, the pressure of 0.60 mole `NH_3` gas in a 3.00 L vessel at `25^@C` is

Options

A. 48.9 atm
B. 4.89 atm
C. 0.489 atm
D. 489 atm

Answer

To find the pressure of 0.60 moles of NH₃ gas in a 3.00 L vessel at 25°C using the ideal gas law, we can follow these steps:

### Step 1: Write down the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature (in Kelvin)

### Step 2: Convert Temperature to Kelvin
The temperature is given as 25°C. To convert this to Kelvin:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 25 + 273.15 = 298.15 \, K \]

### Step 3: Identify the Values
From the problem, we have:
- \( n = 0.60 \, \text{moles} \)
- \( V = 3.00 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \) (universal gas constant)
- \( T = 298.15 \, K \)

### Step 4: Rearrange the Ideal Gas Law to Solve for Pressure
We need to solve for \( P \):
\[ P = \frac{nRT}{V} \]

### Step 5: Substitute the Values into the Equation
Now, substituting the known values into the equation:
\[ P = \frac{(0.60 \, \text{mol}) \times (0.0821 \, \text{L atm/(K mol)}) \times (298.15 \, K)}{3.00 \, \text{L}} \]

### Step 6: Calculate the Pressure
Calculating the numerator:
\[ (0.60) \times (0.0821) \times (298.15) \approx 14.694 \]
Now divide by the volume:
\[ P = \frac{14.694}{3.00} \approx 4.898 \, \text{atm} \]

### Step 7: Round the Result
Rounding to two decimal places:
\[ P \approx 4.89 \, \text{atm} \]

### Conclusion
The pressure of 0.60 moles of NH₃ gas in a 3.00 L vessel at 25°C is approximately **4.89 atm**.

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