Calculate the temperature of 2.0 mol of a gas occupying `3 dm^(3)` at 3.32 bar pressure. (R = 0083 bar `dm^(3) K^(-1) mol^(-1)`)

Answer

To calculate the temperature of the gas, we can use the Ideal Gas Law, which is represented by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in bar)
- \( V \) = volume (in dm³)
- \( n \) = number of moles (in mol)
- \( R \) = ideal gas constant (in bar dm³ K⁻¹ mol⁻¹)
- \( T \) = temperature (in Kelvin)

Given:
- \( n = 2.0 \, \text{mol} \)
- \( V = 3.0 \, \text{dm}^3 \)
- \( P = 3.32 \, \text{bar} \)
- \( R = 0.083 \, \text{bar dm}^3 \, \text{K}^{-1} \, \text{mol}^{-1} \)

### Step 1: Rearrange the Ideal Gas Law to solve for Temperature (T)
From the Ideal Gas Law, we can rearrange the equation to find \( T \):

\[ T = \frac{PV}{nR} \]

### Step 2: Substitute the known values into the equation
Now, we can substitute the values of \( P \), \( V \), \( n \), and \( R \) into the equation:

\[ T = \frac{(3.32 \, \text{bar}) \times (3.0 \, \text{dm}^3)}{(2.0 \, \text{mol}) \times (0.083 \, \text{bar dm}^3 \, \text{K}^{-1} \, \text{mol}^{-1})} \]

### Step 3: Calculate the numerator
Calculate the product of pressure and volume:

\[ 3.32 \, \text{bar} \times 3.0 \, \text{dm}^3 = 9.96 \, \text{bar dm}^3 \]

### Step 4: Calculate the denominator
Calculate the product of number of moles and the gas constant:

\[ 2.0 \, \text{mol} \times 0.083 \, \text{bar dm}^3 \, \text{K}^{-1} \, \text{mol}^{-1} = 0.166 \, \text{bar dm}^3 \, \text{K}^{-1} \]

### Step 5: Divide the numerator by the denominator
Now, divide the results from Step 3 and Step 4 to find \( T \):

\[ T = \frac{9.96 \, \text{bar dm}^3}{0.166 \, \text{bar dm}^3 \, \text{K}^{-1}} \]

\[ T \approx 60.0 \, \text{K} \]

### Final Answer
The temperature of the gas is approximately **60 Kelvin**.

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