\( 4.3 \quad \begin{array}{l}0,61 \mathrm{~kg} \text { of nitrogen at an absolute pressure of } 100 \mathrm{kPa} \text { and with a volume of } \\ 521 \text { litres has a temperature of } 27^{\circ} \mathrm{C}\end{array} \) Calculate the gas constant for nitrogen. [10] TOTAL: 60

1. Write down the ideal gas law in the form that relates pressure, volume, mass, the specific gas constant, and temperature: \[ pV = mRT \] where \( p \) is the pressure, \( V \) is the volume, \( m \) is the mass, \( R \) is the specific gas constant for nitrogen, and \( T \) is the absolute temperature. 2. Rearranging for \( R \) gives: \[ R = \frac{pV}{mT} \] 3. Convert all given quantities to SI units: - Pressure: \( p = 100\,\text{kPa} = 100\,000\,\text{Pa} \) - Volume: \( V = 521\,\text{L} = \frac{521}{1000}\,\text{m}^3 = 0.521\,\text{m}^3 \) - Temperature: \( T = 27^\circ\text{C} = 27 + 273.15 = 300.15\,\text{K} \) - Mass is already in kilograms: \( m = 0.61\,\text{kg} \) 4. Substitute the converted values into the formula for \( R \): \[ R = \frac{100\,000 \times 0.521}{0.61 \times 300.15} \] 5. Calculate the numerator: \[ 100\,000 \times 0.521 = 52\,100\,\text{Pa}\cdot\text{m}^3 \] 6. Calculate the denominator: \[ 0.61 \times 300.15 \approx 183.0915\,\text{kg}\cdot\text{K} \] 7. Now, compute \( R \): \[ R \approx \frac{52\,100}{183.0915} \approx 284.6\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \] Thus, the gas constant for nitrogen is approximately \[ R \approx 285\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}. \]