What is the starting temperature of 0.250L of

gas when cooled to 306K and a volume of

0.320L?

How is it done

239K

Explanation:

Data obtained from the question is given below:

V1 = 0.250L

T1 = ?

T2 = 306K

V2 = 0.320L

Now, we can easily find the starting temperature by doing the following:

V1 / T1 = V2/T2

0.250 / T1 = 0.320/306

Cross multiply to express in linear form

T1 x 0.320 = 0.250 x 306

Divide both side 0.320

T1 = (0.250 x 306) / 0.320

T1 = 239K