Question

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

Answer 1

Answer:

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:

           $\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

Let us take the units to the appropriate one;

      -550°C  = 273 + (-550) = -277K

Input the variables;

             $\frac{760 x 3.9}{273} = \frac{808 x V_{2} }{-277}$

        V₂  = 3.72L