Pedro23padron
Pedro23padron Pedro23padron
  • 16-03-2021
  • Chemistry
contestada

A container of 56mL is at STP, The pressure is increased to 3.6atm and the volume to 162 mL.
What is the new temperature?

Respuesta :

jcherry99
jcherry99 jcherry99
  • 16-03-2021

Answer:

T2 = 2843.1   oK. This is a huge temperature.  Check it for errors.

Explanation:

Remark

This is the same question as the other one I've answered. Only the numbers have been altered.

Givens

v1 = 56 mL

P1 = 1 atm

T1 = 273o K

v2 = 162

P2 = 3.6 atm

T2 = ?

Formula

Vi * P1 / T1 = V2 * P2/T2

Solution

Rearrange the formula so T2 is on the left

T2 = V2 P2 * T1 / (V1 * P1)          Now just put the numbers in.

T2 = 162 * 3.6* 273 / (56 *1)

T2  =  159213.6/56

T2 = 2843.1                  

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