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

1 Answer
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,

V/T=k where, k is a constant

So,putting V=580 ml and T=17+273=290K we get,

580/290=k

Now,if new temperature would be T_1 then,

700/(T_1)=k=580/290

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