Question

A gas is collected at 22.0 ˚C and 745.0 mm Hg. When the temperature is changed to 0 ˚C, what is the resulting pressure?

Answer 1

Explanation:

Problem:

A gas collected at 22.0°c and 745.0 mm Hg. When the temperature is changed to 0°c.What is the resulting pressure.

Given Data:

P₁ (initial pressure) = 745.0 mm Hg

T₁ (initial temperature) = 22°c + 273 = 295 K

P₂ (final pressure)  =  unknown

T₂ (final temperature) = 0°c + 273 = 273 K

Solution:

We can use the formula of Gay-Lussac's Law which is P₁/T₁ = P₂/T₂ and to find P₂ we can use the formula P₂ = P₁T₂/T₁

P₂ = P₁T₂/T₁

    = 745.0 mm Hg × 273 K  /  295 K

   = 689.44 mm Hg

Answer:

the resulting pressure would be 689.44 mm Hg

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Answer 2

Answer:

689.44

Explanation:

here the question is

a gas collected at 22 degrees Celsius in 745 mm of Hg. when the temperature is changed to 0 degrees Celsius what is the resultant pressure

so here formula is

p1/t1=p2/t2

p1=745 , p2=?

t1=22+273=295 , t2=0+273=273

=745/295=p2/273

=745 into 273/295

=203385/295

=689.44

so the value of p2=689.44

so hence we proved the resultant pressure