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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

  1. At what temperature will 0.654 mol of neon gas occupy 12.30 L at 1.95 atm?*1 point264 K314 K389 K447 K
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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.

This problem has been solved

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