A 3.90L sample of gas at STP is cooled to -55oC at 808mmHg. What is the new volume?

3.72L

Explanation:

Given parameters:

Initial volume V₁ = 3.9L

Condition  = STP

Final temperature T₂ = -550°C

Final pressure P₂  = 880mmHg

Unknown:

Final volume V₂  = ?

Solution.

At standard temperature and pressure(STP), the:

           Pressure  = 1atm   = 760mmHg

            Temperature  = 273K

Therefore, P₁  = 760mmHg

                  T₁  = 273K

The general gas law, is best to solve this problem. It is mathematically given as:

Let us take the units to the appropriate one;

      -550°C  = 273 + (-550) = -277K

Input the variables;

        V₂  = 3.72L

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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