79.6k views
4 votes
When the pressure that a gas exerts

on a sealed container changes
from atm to 1.80 atm, the
temperature changes from 86.0°C to
30.0°C.

User Malkie
by
7.9k points

2 Answers

4 votes

Answer:

2.13

Step-by-step explanation:

User Robert Kock
by
7.8k points
4 votes

Answer:

The pressure changes from 2.13 atm to 1.80 atm.

Step-by-step explanation:

Given data:

Initial pressure = ?

Final pressure = 1.80 atm

Initial temperature = 86.0°C (86.0 + 273 = 359 K)

Final temperature = 30.0°C (30+273 =303 K)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

P₁ = P₂T₁ /T₂

P₁ = 1.80 atm × 359 K / 303 K

P₁ = 646.2 atm. K /303 K

P₁ = 2.13 atm

The pressure changes from 2.13 atm to 1.80 atm.

User Jkigel
by
8.0k points
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