At what temperature (in K) would 2.10 mol of N2 gas have a pressure of 1.25 atm in a 22.1 L tank? (R = 0.0821 L·atm/mol·K.
Question
At what temperature (in K) would 2.10 mol of N2 gas have a pressure of 1.25 atm in a 22.1 L tank?
(R = 0.0821 L·atm/mol·K)
Solution
To solve this problem, we can use the ideal gas law equation, which is:
PV = nRT
Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature
We are asked to find the temperature (T), so we rearrange the equation to solve for T:
T = PV / nR
Now we can substitute the given values into the equation:
T = (1.25 atm * 22.1 L) / (2.10 mol * 0.0821 L·atm/mol·K)
Now, calculate the value:
T = 27.75 / 0.17241
T = 160.9 K
So, the temperature of the gas would be approximately 161 K.
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