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What pressure is exerted by 0.344 mol of #N_2# in a 9.73 L steel container at 105.4°C?

Answer 1

Avery Adams

This is a clear application of the Ideal Gas equation..........I get approx. #1*atm#.............

#P=(nRT)/V=(0.344*cancel(mol)xx0.0821*(cancelL*atm)/(cancelK*cancel(mol))xx378.5*cancelK)/(9.73*cancelL)#

#~=1*atm#

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Answer 2

Noah Aldrich

Use the ideal gas law: (PV = nRT) Given:

(n = 0.344 , \text{mol})
(V = 9.73 , \text{L})
(T = 105.4°C) (convert to Kelvin by adding 273.15)
(R = 0.0821 , \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))

Plug in values and solve for (P): [P = \frac{nRT}{V}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.