The pressure of one mole of an ideal gas of volume 10$^-2$ m$^3$ at a temperature of 27°c is [molar gas constant = 8.3 J mol$^{-1}$ k$^{-1}$]

Analyze the question: We need the pressure of 1 mole of an ideal gas with V = 10⁻² m³ at 27°C, using R = 8.3 J mol⁻¹ K⁻¹.

Concept: The ideal gas equation is PV = nRT, so P = nRT / V.

Solution:

1. Convert the temperature to kelvin: T = 27 + 273 = 300 K

2. Substitute the values: P = (1 × 8.3 × 300) / 10⁻²

3. Calculate the numerator: 8.3 × 300 = 2490 J

4. Divide by the volume: P = 2490 / 0.01 = 249,000 N/m² P = 2.49 × 10⁵ N m⁻²

Answer: This matches option C (2.

Option A has the right digits but the wrong power of ten (10⁴). Option D has the wrong unit (Nm⁻¹ instead of Nm⁻²) and the wrong value.

Tip: Always convert temperature to kelvin first (add 273). Then watch the power of ten, since dividing by 10⁻² is the same as multiplying by 100. Here 2490 × 100 = 249,000. Also check that the unit is N m⁻² (pascal) for pressure.