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At what temperature would the volume of gas be doubled, if the pressure at the same time increases from 700-800 mm, the gas being at 0°c initially ?

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The situation described can be solved using the combined gas law, which states:

(P1 x V1) / (T1) = (P2 x V2) / (T2)

Where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, respectively. P2, V2, and T2 are the final pressure, volume, and temperature of the gas, respectively.

To solve for the final temperature (T2), we can rearrange the equation as:

T2 = (P2 x V2 x T1) / (P1 x V1)

Let's plug in the given values:

P1 = 700 mm

P2 = 800 mm

V1 = V2 (since the volume is doubled)

T1 = 0°C + 273.15 = 273.15 K (converting to Kelvin)

So, we have:

T2 = (800 mm x 2V1 x 273.15 K) / (700 mm x V1)

T2 = (1600/7) x 273.15 K

T2 = 392.2 K

Therefore, the final temperature at which the volume of gas would be doubled, if the pressure increases from 700-800 mm and the gas is initially at 0°C, is approximately 392.2 Kelvin (which is approximately 119°C or 246°F).

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