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Sedaia [141]
3 years ago
11

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

Chemistry
1 answer:
notsponge [240]3 years ago
5 0

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

           

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