Question

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

Answer 1

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`