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6 May, 12:19

What is the starting temperature of 0.250L of

gas when cooled to 306K and a volume of

0.320L?

How is it done

+3
Answers (1)
  1. 6 May, 15:21
    0
    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
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