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ruslelena [56]
3 years ago
8

  If the temperature is kept constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one-th

ird of its original value?
Chemistry
1 answer:
joja [24]3 years ago
3 0
Using the ideal gas equation pV = nRT, where R is the ideal gas constant, we can show that p=nRT/V. Since n, R and T are all constants, p2/p1 = V1/V2 where p1 and p2 are the start and final pressures respectively and V1 and V2 are the start and final volumes respectively. For if p1 = 3*p2, the pressure would have fallen to one third of its original value, and it follows that V2 = 3*V1. Therefore, for the pressure to fall to a third of its original value, the volume must increase by a factor of 3.
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The vapor pressure of the solution is 23.636 torr

Explanation:

P_{solution} = X_{solvent}*P_{solvent}

Where;

P_{solution is the vapor pressure of the solution

X_{solvent is the mole fraction of the solvent

P_{solvent is the vapor pressure of the pure solvent

Thus,

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Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

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