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

answer; i believe that the correct answer is (a) potassium(k)

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