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

To calculate the gas constant \( R \) for nitrogen, we use the ideal gas equation, which relates pressure \( P \), volume \( V \), temperature \( T \), and the amount of gas \( n \): \[ PV = nRT \] First, we need to convert the standard conditions to match the units. Here, the pressure is given as \( P = 100 \, \text{kPa} = 100,000 \, \text{Pa} \) and the volume is \( V = 521 \, \text{litres} = 0.521 \, \text{m}^3 \). The temperature must also be converted from Celsius to Kelvin: \[ T = 27^{\circ}C + 273.15 = 300.15 \, K \] Next, we can determine the amount of nitrogen gas \( n \) using its mass. The molecular weight of nitrogen \( N_2 \) is approximately \( 28 \, \text{g/mol} \), or \( 0.028 \, \text{kg/mol} \). Thus, the number of moles \( n \) can be calculated as follows: \[ n = \frac{0.61 \, \text{kg}}{0.028 \, \text{kg/mol}} \approx 21.79 \, \text{mol} \] Now we can rearrange the ideal gas equation to find \( R \): \[ R = \frac{PV}{nT} \] Substituting the values we calculated: \[ R = \frac{(100,000 \, \text{Pa})(0.521 \, \text{m}^3)}{(21.79 \, \text{mol})(300.15 \, \text{K})} \] Calculating \( R \): \[ R \approx \frac{52,100}{6,548.07} \approx 7.95 \, \text{J/(mol K)} \] So, the gas constant for nitrogen in this particular scenario is approximately \( 7.95 \, \text{J/(mol K)} \).