The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023

<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
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50 cl is 500 millilitres
55nm is 5.5 × 10^-6 centimetre
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>