Question

A `42.3*mL` volume of gas at `371.3*K` is cooled to `351.7*K`. What is the new volume?

Answer 1

We get approx. `V_2<=50*mL`

Explanation:

We use old `"Charles' Law"` to give `VpropT` for a given quantity of gas...........

Thus `V=kT`, and solving for `k` in each scenario.....

`V_1/T_1=V_2/T_2`, ad of course `T` is quoted in `"degrees Kelvin"`.

And so we solve for `V_2=V_1/T_1xxT_2=(42.3*mLxx351.7*K)/(371.3*K)`

`=??*mL........`