What is the volume of a gas at standard temperature if the gas occupies 3.65L at 25 Celsius?
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The standard conditions in chemistry refer to a temperature of 273 K or 0 Celsius and a pressure if 1 atm.
Here the volume of the gas at 25 Celsius is 3.65 l. Use the Ideal Gas Law which gives P*V = n*R*T where P is the pressure, V is the volume, n is the number of moles of the gas, R is a constant and T is the temperature.
Here we know V = 3.65 at 298 K
P*3.65 = n*R*298
=> 3.65/298 = n*R/P = a constant which we can denote by k
At the standard temperature of 273 K, let the volume be V
V/273 = k = 3.65/298
=> V = 3.65*273/298
=> V = 3.343
The volume of the gas at standard temperature is 3.343 l
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This problem could be solve using Charles' Law:
V1/T1 = V2/T2
Let the volume to be determined be x.
Since the temperatures are given in Kelvin scale, we'll have to convert 25 Celsius into Kelvin.
T1 = 25 + 273
T1 = 298 K
Now, we'll replace all we know in Charles' Law:
3.65L/298K = x/273K
273 K is the standard temperature.
We'll cross multiply and we'll get:
3.65L*273K = x*298K
x = 3.65L*273K/298K
x = 3.3437L
The requested volume of gas, at standard temperature, is of 3.3437L.
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