1)If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to 550 C, what will the new volume of the balloon be?

V₂ = 70.80 L

Explanation:

Given data:

Initial volume of He = 45 L

Initial temperature = 250 °C

Final temperature = 550 °C

Final volume = ?

Solution:

Initial temperature = 250 °C (250+273.15 K= 523.15 K)

Final temperature = 550 °C (550 + 273.15 k= 823.15 K)

According to Charles Law.,

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 45 L × 823.15  K / 523.15 K

V₂ = 37041.75 L.K / 523.15 K

V₂ = 70.80 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.

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