What volume would be occupied by 115 g of nitrogen gas in at 8.2 atm pressure and 65oC?

Question

BlackTom AI · Accepted Answer

To determine the volume, we’ll use the ideal gas law in the form PV = nRT.
First, convert the given mass of nitrogen gas to moles. Nitrogen gas is N2 with a molar mass of about 28.0 g/mol, so n = 115 g / 28.0 g/mol ≈ 4.11 mol.
Next, convert the temperature to kelvin: 65°C + 273.15 ≈ 338 K.
Now plug into PV = nRT: V = nRT / P = (4.11 mol)(0.08206 L·atm/(mol·K))(338 K) / (8.2 atm).
Carrying out the calculation: (0.08206 × 338) ≈ 27.75; (27.75 × 4.11) ≈ 114.0; 114.0 / 8.2 ≈ 13.9 L, which rounds to about 14 L.
Option-by-option analysis:
- 14 L: This is consistent with the calculation above, so this choice is correct given the data.
- 2.7 L: This is far too small for ~4.1 moles at 338 K and 8.2 atm; it would imply a much smaller volume than the gas actually occupies under these conditions.
- 0.4 L: An even more extreme underestimation; such a small volume would correspond to extremely high pressure or far fewer moles, neither of which matches the given values.
- 0.8 L: Similar to the previous two, this would require conditions that compress the gas far beyond the provided P, n, and T; it contradicts the calculation.
In summary, the computed volume around 13.9–14 L aligns with the ideal gas law using the provided n, T, and P, whereas the other options reflect volumes that do not fit these conditions.

What volume would be occupied by 115 g of nitrogen gas in at 8.2 atm pressure and 65oC?

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