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A balloon of air now occupies 5.0 L, at 25 celcius and 1.00 atm. What temperature was it initially, if it occupied 4.0 L and was in a very cold freezer with a pressure of 0.92 atm?
combined gas laws
A. 219k B. 298k C.325k D.425k

Respuesta :

Ajiduah1

Answer: A. 219K

Explanation:

Using the Combined Gas Law; P1 V1/T1 = P2V2/T2

Where P1 = 0.92 atm, P2 = 1.00 atm

           V1 =4.0 L,   V2 = 5.0 L

           T1 =  ?,  T2 = 25 celcius = 298K

We therefore make T1 the subject of the formular;

T1 = P1 . V1 .T2 / P2. V2

    = 0.92atm x 4.0L x 298K / 1.00 atm x 5.0 L

    = 1096.64 /   5

    = 219.328 approx  219K

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