If a quantity of heat 1163 . 4 J is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (Given R = 8 . 31 J mole - 1 K - 1 )

Question

If a quantity of heat 1163 . 4 J is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (Given R = 8 . 31 J mole - 1 K - 1 ), 54 K, 28 K, 65 K, 40 K

MARKS App · Accepted Answer

Since nitrogen is diatomic gas C ⁡ P ⁡ = 7 2 R ⁡ Heat supplied Q = n C p Δ T where n is the number of moles, C P is the molar specific hence at constant pressure, ΔT is the rise in temperature. ∴ 1 1 6 3 · 4 = 1 7 2 × 8 · 3 1 Δ T ⁡ Δ T ⁡ = 1 1 6 3 · 4 × 2 7 × 8 · 3 1 = 4 0 K

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