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A container of oxygen gas is at STP. If this sample is put into an oven at 280 C, what would its pressure be, in atmospheres?
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Answers ( )
Answer:
Explanation:
We are concerned with the variables of temperature and pressure, so we use Gay-Lussac’s Law, which states the temperature of a gas is directly proportional to the pressure. The formula is:
We know that the container of gas begins at standard temperature and pressure (STP). This is 1 atmosphere of pressure and 273 Kelvin.
We know the gas is put into an oven at 280 degrees Celsius. We can convert this to Kelvin.
We are solving for the new pressure, so we must isolate the variable P₂. It is being divided by 553.15 Kelvin. The inverse of division is multiplication, so we multiply both sides by 553.15 K
The units of Kelvin cancel.
Rounded to the nearest hundredth:
The new pressure is approximately 2.03 atmospheres.
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial pressure (P1) = 1atm
Initial temperature (T1) = 0°C = 0°C + 273 = 273K
Final temperature (T2) = 280°C = 280°C + 273 = 553K
Final pressure (P2) =…?
Step 2:
Determination of the new pressure of the gas.
Since the volume of the gas is constant, the following equation:
P1/T1 = P2/T2
will be used to obtain the pressure. This is illustrated below:
P1/T1 = P2/T2
1/273 = P2 / 553
Cross multiply
273x P2 = 553
Divide both side by 273
P2 = 553/273
P2 = 2.03atm
Therefore, the new pressure of the gas will be 2.03atm