the pressure in a sealed plastic container is 108 kPa at 41 degrees Celsius. What is the pressure when the temperature drops to
1 answer:
<u>Answer:</u> The new pressure will be 101.46 kPa.
<u>Explanation:</u>
To calculate the new pressure, we use the equation given by Gay-Lussac Law. This law states that pressure is directly proportional to the temperature of the gas at constant volume.
The equation given by this law is:
where,
are initial pressure and temperature.
are final pressure and temperature.
We are given:
By using conversion factor:
Putting values in above equation, we get:
Hence, the new pressure will be 101.46 kPa.
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Answer:
Value of equilibrium constant is 0.0888
Explanation:
Both and
are gaseous. Hence equilibrium constant depends upon partial pressures of
and
.
Initially no and
were present.
Hence mole fraction of and
at equilibrium can be calculated from coefficient of
and
in balanced equation.
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Let's assume both and
behaves ideally.
Therefore partial pressure of ,
and P_{CO_{2}}= x_{CO_{2}}.P_{total}
Where x represents mole fraction
So,
So,