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`, we will use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as \( 25°C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 25 + 273.15 = 298.15 \, K \]

### Step 2: Identify the values for the ideal gas law
From the problem, we have:
- \( n = 0.60 \, \text{moles} \)
- \( V = 3.00 \, \text{L} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 298.15 \, K \)

### Step 3: Rearrange the ideal gas law to solve for pressure \( P \)
We need to isolate \( P \) in the ideal gas law equation:

\[ P = \frac{nRT}{V} \]

### Step 4: Substitute the known values into the equation
Now, we substitute the values we have 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 5: Calculate the pressure
Now we perform the calculation:

1. Calculate the numerator:
\[ (0.60) \times (0.0821) \times (298.15) \approx 14.696 \, \text{L·atm} \]

2. Divide by the volume:
\[ P = \frac{14.696}{3.00} \approx 4.8987 \, \text{atm} \]

### Step 6: Round the result
Rounding to two decimal places, we find:
\[ P \approx 4.90 \, \text{atm} \]

### Final Answer
The pressure of 0.60 moles of `NH3` gas in a 3.00 L vessel at `25°C` is approximately **4.90 atm**.

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