40. At what temperature will 0.654 mol of neon gas occupy 12.30 L at 1.95 atm?*1 point264 K314 K389 K447 K
Question
40. At what temperature will 0.654 mol of neon gas occupy 12.30 L at 1.95 atm?
1 point
- 264 K
- 314 K
- 389 K
- 447 K
Solution
To solve this problem, we can use the ideal gas law equation, which is PV = nRT.
Here, P = pressure = 1.95 atm V = volume = 12.30 L n = number of moles = 0.654 mol R = ideal gas constant = 0.0821 L.atm/K.mol (since the pressure is given in atm and volume in L) T = temperature = ?
We need to find the temperature, so we rearrange the equation to solve for T:
T = PV / nR
Substitute the given values into the equation:
T = (1.95 atm * 12.30 L) / (0.654 mol * 0.0821 L.atm/K.mol)
Calculate the above expression to find the temperature. The answer should be in Kelvin.
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