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29 votes
29 votes
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?

User Shan Shafiq
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1 Answer

28 votes
28 votes

Answer:

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

Step-by-step 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

User Jene
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