Question 2 of 7intermediate🔧 ApplyNumerical3 marks

What amount of heat must be supplied to 2.0×10−2 kg2.0 \times 10^{-2} \text{ kg} of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2=28N_2 = 28; R=8.3 J mol−1 K−1R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}).

Correct Answer

Given: Mass of nitrogen, m = 2.0×102.0 \times 10⁻² kg; Temperature rise, ΔT = 45 °C = 45 K; Molecular mass of N₂ = 28 g/mol; R = 8.3 J mol⁻¹ K⁻¹

Step 1: Calculate number of moles n = Mass / Molar mass = (2.0×102.0 \times 10⁻² kg) / (28×1028 \times 10⁻³ kg/mol) = 0.02/0.028 = 5/7 mol

Step 2: Determine molar specific heat at constant pressure (Cp) Nitrogen (N₂) is a diatomic gas. At room temperature, it has 5 degrees of freedom (3 translational + 2 rotational). Cv = (f/2)R = (5/2)R Cp = Cv + R = (5/2)R + R = (7/2)R = (7/2) × 8.3 = 29.05 J mol⁻¹ K⁻¹

Step 3: Calculate heat supplied Q = n × Cp × ΔT Q = (5/7) × (7/2) × 8.3×458.3 \times 45 Q = (5/2) × 8.3×458.3 \times 45 Q = 2.5×8.32.5 \times 8.3 × 45 Q = 933.75 J ≈\approx 934 J

Therefore, the amount of heat that must be supplied is 934 J.

Exercise: EXERCISES | Q: 11.2 | (Page Page 17)
For More Understanding

Explanation

This question tests understanding of molar specific heat capacities for diatomic gases. At room temperature, nitrogen (N₂) behaves as a diatomic gas with 5 degrees of freedom. The key formula is Q = nCpΔT for constant pressure process. Students must remember that Cp = Cv + R, and for diatomic gases, Cv = (5/2)R, giving Cp = (7/2)R.

Solution Steps

  1. Step 1: Calculate number of moles using n = m/M

  2. Step 2: Determine Cp for diatomic gas: Cp = (7/2)R

  3. Step 3: Apply Q = nCpΔT to find heat supplied

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