A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and
32 C.

1255.4L

Explanation:

Given parameters:

P₁  = 928kpa

T₁  = 129°C

V₁  = 569L

P₂ = 319kpa

T₂  = 32°C

Unknown:

V₂  = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

 take the units to the appropriate ones;

             kpa to atm,  °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁  = 928kpa gives 9.16atm

T₂  = 32°C gives 273 + 32  = 305K

T₁  = 129°C gives 129 + 273  = 402K

Input the values in the equation and solve for V₂;

       V₂   = 1255.4L

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.