# To what calsius temperature must 580 ml of oxygen at 17°C be raised to increase its volume 700 mL?

Mar 22, 2018

For doing the mentioned process under isobaric condition for constant mass of ${O}_{2}$ we can apply Charle's Law.

So,from Charle's Law we know,

$\frac{V}{T} = k$ where, $k$ is a constant

So,putting $V = 580 m l$ and $T = 17 + 273 = 290 K$ we get,

$\frac{580}{290} = k$

Now,if new temperature would be ${T}_{1}$ then,

$\frac{700}{{T}_{1}} = k = \frac{580}{290}$

So, T_1=(700×290)/580=350K=(350-273)^@C=77^@C