2 years ago

What volume change occurs to a 400.0 mL gas sample as the temperature increases from 22.0 °C to 30.0°C?

1 answer:

4 0

V₁ = 400.0 mL

T₁ = 22.0ºC + 273 = 295 K

V₂ = ?

T₂ = 30.0ºC + 273 = 303 K

V₁ / T₁ = V₂ / T₂

400.0 / 295 = V₂ / 303

295 x V₂ = 400.0 x 303

295 V₂ = 121200

V₂ = 121200 / 295

V₂ = 410.84 mL

hope this helps!

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