At what temperature, would the volume of a gas to be doubled. If pressure at the same time increases
from 700 to 800mm, the gas being at
initially
at 0°C

Explanation:

P1V1 = nRT1

P2V2 = nRT2

Divide one by the other:

P1V1/P2V2 = nRT1/nRT2

From which:

P1V1/P2V2 = T1/T2

(Or P1V1 = P2V2 under isothermal conditions)

Inverting and isolating T2 (final temp)

(P2V2/P1V1)T1 = T2 (Temp in K).

Now P1/P2 = 1

V1/V2 = 1/2

T1 = 273 K, the initial temp.

Therefore, inserting these values into above:

2 x 273 K = T2 = 546 K, or 273 C.

Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.

From the ideal gas equation:

V = nRT/P or at constant pressure:

V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.

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.

the product of the pressure and volume is also constant