A rigid tank of 300 litre capacity contains 3 kg of oxygen (molar mass = 32 kg/kmol) at 25°C. If oxygen behaves as an ideal gas, the pressure (in kPa) inside the tank is ......... (rounded off to two decimal places).
Use: Universal gas constant \( R_u = 8.314 \, {kJ/kmol-K} \)
Use: Universal gas constant \( R_u = 8.314 \, {kJ/kmol-K} \)
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Solution and Explanation
- Volume \( V = 300 \, {litre} = 0.3 \, {m}^3 \)
- Mass of oxygen \( m = 3 \, {kg} \)
- Molar mass of oxygen \( M = 32 \, {kg/kmol} \)
- Temperature \( T = 25^\circ {C} = 298.15 \, {K} \)
First, calculate the number of moles \( n \) of oxygen:
\[ n = \frac{m}{M} = \frac{3}{32} = 0.09375 \, {kmol} \] Now, using the ideal gas law:
\[ PV = nRT \] Rearrange to find the pressure \( P \):
\[ P = \frac{nRT}{V} \] Substitute the values:
\[ P = \frac{0.09375 \times 8.314 \times 298.15}{0.3} = 773.07 \, {kPa} \] Thus, the pressure inside the tank is approximately \( 773.07 \, {kPa} \), which lies between 773 and 776 kPa.
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