In a 10.0 L vessel at 100.0 oC, 10.0 grams of an unknown gas exert a pressure of 1.13 atm. What is the gas?
PV = nRT R = 0.0821 (L·atm)/(mol·K)

Question

In a 10.0 L vessel at 100.0 oC, 10.0 grams of an unknown gas exert a pressure of 1.13 atm.  What is the gas?
PV = nRT            R = 0.0821 (L·atm)/(mol·K)

BlackTom AI · Accepted Answer

PV = nRT, so find n then M = mass/n.

1) Relevant knowledge points and concepts
- Ideal gas law: PV = nRT relates pressure (P), volume (V), moles (n), gas constant (R), and temperature (T in K).
- Molar mass M = mass (g) / amount (mol).
- Temperature must be in Kelvin: T(K) = T(°C) + 273.15.

2) Complete reasoning process step by step
- Given: P = 1.13 atm, V = 10.0 L, mass = 10.0 g, T = 100.0 °C = 100.0 + 273.15 = 373.15 K, R = 0.0821 L·atm·mol–1·K–1.
- Solve PV = nRT for n: n = PV / (R T).

3) Calculations with intermediate steps
- PV = (1.13 atm)(10.0 L) = 11.3 L·atm.
- R·T = 0.0821 L·atm·mol–1·K–1 × 373.15 K = 30.635615 L·atm·mol–1.
- n = 11.3 / 30.635615 = 0.3690 mol (rounded).
- Molar mass M = mass / n = 10.0 g / 0.3690 mol = 27.10 g·mol–1 ≈ 27 g·mol–1.

4) Why the correct answer is correct
The calculated molar mass is about 27.1 g·mol–1. The molecular formula HCN (hydrogen cyanide) has atomic masses H = 1, C = 12, N = 14, giving M = 1 + 12 + 14 = 27 g·mol–1, matching the computed value. Therefore the gas is HCN.

5) Why the other options are incorrect
- NH3: molar mass = 14 (N) + 3×1 (H) = 17 g·mol–1, not 27.
- NO (option lists 39 g·mol–1): regardless of the erroneous 39 value, true NO molar mass is 14 + 16 = 30 g·mol–1, not 27.
- NO2: molar mass = 14 + 2×16 = 46 g·mol–1, not 27.
None match the calculated ~27 g·mol–1.

6) Conclusion
The computed molar mass (≈27.1 g·mol–1) corresponds to HCN, so the correct answer is: The gas is HCN and the molar mass is 27 g·mol–1. This matches the provided answer.

In a 10.0 L vessel at 100.0 oC, 10.0 grams of an unknown gas exert a pressure of 1.13 atm. What is the gas? PV = nRT R = 0.0821 (L·atm)/(mol·K)单项选择题

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