A sample of gas has a volume of 42 L at a pressure of 200 kPa and a temperature of 30^ * C . What would be the volume, if the pressure were increased to 250 kPa and the temperature were decreased to 5^ * C ?

30 L

Explanation:

Ideal gas law:

PV = nRT

Rearranging:

PV / T = nR

Since n and R are constant:

P₁ V₁ / T₁ = P₂ V₂ / T₂

Plugging in values and solving:

(200 kPa) (42 L) / (30 + 273.15 K) = (250 kPa) V / (5 + 273.15 K)

V = 30.8 L

Rounded to one significant figure, the new volume is 30 L.

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= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation