Question

At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0 °C?

Answer 1

V₁ = 400.0 mL

T₁ = 225.0ºC + 273 = 498 K

V₂ = ?

T₂ = 127.0ºC + 273 = 400 K

V₁ / T₁ = V₂ / T₂

400.0 / 498 = V₂ / 400

498 x V₂ = 400 x 400

498 x V₂ = 160000

V₂ = 160000 / 498

V₂ = 321.28 mL

hope this helps!

Answer 2

The volume of the gas at 127.0 °C is 321.28 ml.

What is volume?

The volume of any substance is the total space it occupies.

Given data is

V₁ is 400.0 mL

T₁  is 225.0ºC + 273 = 498 K

T₂ = 127.0ºC + 273 = 400 K

To find V₂

By Ideal gas law

$\rm \dfrac{V_1}{T_1 } =\dfrac{V_2}{T_2}$

$\rm \dfrac{400.0}{498 } =\dfrac{V_2}{400}\\\\\\ \dfrac{160000}{498}= 321.28 ml$

Thus, the volume of the gas at 127.0 °C is 321.28 ml.

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