The compressibility factor `(Z=PV//nRT)` for `N_(2)` at `223 K` and `81.06 MPa` is `1.95`, and at `373 K` and `20.265 MPa`, it is `1.10`. A certain mass of `N_(2)` occupies a volume of `1.0 dm^(3)` at `223 K` and `81.06 MPa`. Calculate the volume occupied by the same quantity of `N_(2)` at `373 K` and `20.265 MPa`.

Options

A. `3.774 dm^(3)`
B. `2.77 dm^(3)`
C. `5.07 dm^(3)`
D. `9.30 dm^(3)`

Explanation

To solve the problem, we will use the compressibility factor formula \( Z = \frac{PV}{nRT} \) and follow these steps:

### Step 1: Calculate the number of moles of \( N_2 \) at the first condition (223 K, 81.06 MPa).
We can rearrange the formula to find \( n \):
\[ n = \frac{PV}{ZRT} \]
Given:
- \( P = 81.06 \, \text{MPa} = 81.06 \times 10^6 \, \text{Pa} \)
- \( V = 1.0 \, \text{dm}^3 = 1.0 \times 10^{-3} \, \text{m}^3 \)
- \( Z = 1.95 \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 223 \, \text{K} \)

Now substituting the values into the equation:
\[ n = \frac{(81.06 \times 10^6) \times (1.0 \times 10^{-3})}{1.95 \times 8.314 \times 223} \]

Calculating the numerator:
\[ 81.06 \times 10^6 \times 1.0 \times 10^{-3} = 81060 \, \text{Pa m}^3 \]

Calculating the denominator:
\[ 1.95 \times 8.314 \times 223 \approx 3,702.57 \, \text{J/mol} \]

Now, calculate \( n \):
\[ n \approx \frac{81060}{3702.57} \approx 21.89 \, \text{mol} \]

### Step 2: Calculate the volume occupied by the same quantity of \( N_2 \) at the second condition (373 K, 20.265 MPa).
We will again use the compressibility factor formula rearranged for volume \( V \):
\[ V = \frac{ZnRT}{P} \]
Given for the second condition:
- \( Z = 1.10 \)
- \( T = 373 \, \text{K} \)
- \( P = 20.265 \, \text{MPa} = 20.265 \times 10^6 \, \text{Pa} \)

Now substituting the values into the equation:
\[ V = \frac{(1.10) \times (21.89) \times (8.314) \times (373)}{20.265 \times 10^6} \]

Calculating the numerator:
\[ 1.10 \times 21.89 \times 8.314 \times 373 \approx 1.10 \times 21.89 \times 3100.762 \approx 1.10 \times 67809.39 \approx 74589.29 \]

Calculating the denominator:
\[ 20.265 \times 10^6 = 20265000 \, \text{Pa} \]

Now, calculate \( V \):
\[ V \approx \frac{74589.29}{20265000} \approx 0.00368 \, \text{m}^3 \]
Converting to dm³:
\[ V \approx 3.68 \, \text{dm}^3 \]

### Final Answer:
The volume occupied by the same quantity of \( N_2 \) at 373 K and 20.265 MPa is approximately \( 3.68 \, \text{dm}^3 \).

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