An unknown volume of gas has a pressure of 0.50 atm and temperature of 325 k. if the pressure is raised to 1.2 atm and the temperature decreased to 320 k, giving a new volume of 48 l, what was the initial volume?
a. 65 l
b. 87 l
c. 120 l
d. 140 l

Answer : The correct option is, (C) 120 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

where,

= initial pressure of gas = 0.50 atm

= final pressure of gas = 1.2 atm

= initial volume of gas = ?

= final volume of gas = 48 L

= initial temperature of gas = 325 K

= final temperature of gas = 320 K

Now put all the given values in the above equation, we get:

Therefore, the initial volume was 120 L.

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.

i think it's c. i'm not sure but i think so.