A gas at STP occupies 0.055 L of space. If the pressure changes to 2.8 atm and the temperature increases to 195°C, what is the new
volume?

0.034 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 0.055 L

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = stp = 273 K

Final pressure (P₂) = 2.8 atm

Final temperature (T₂) = 195 °C = 195 °C + 273 = 468 K

Final volume (V₂) =?

The final volume of the gas can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

1 × 0.055 / 273 = 2.8 × V₂ / 468

0.055 / 273 = 2.8 × V₂ / 468

Cross multiply

273 × 2.8 × V₂ = 0.055 × 468

764.4 × V₂ = 25.74

Divide both side by 764.4

V₂ = 25.74 / 764.4

V₂ = 0.034 L

Therefore, the new volume of the gas is 0.034 L.

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