JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Chemical Thermodynamics

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2 5^{\circ} C, 50 g

of iron reacts with HCl to form

FeCl_{2}

. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is _____ J. (Round off to the nearest integer).
[Given:

R = 8.314 J mol^{- 1} K^{- 1}

. Assume hydrogen is an ideal gas]
[Atomic mass of Fe is 55.85 u]

Answer: 2218

Solution

The reaction is:

Fe + 2 HCl FeCl X_{2} + H X_{2}

Moles of Fe =

\frac{50}{55.85} = 0.895

mol

From stoichiometry, 1 mole of Fe produces 1 mole of

H X_{2}

, so moles of

H X_{2}

= 0.895 mol

For expansion against constant external pressure:

W = - P_{e x t} Δ V = - P_{e x t} (V_{f} - V_{i}) \approx - P_{e x t} V_{f}

(since

V_{i} \approx 0

)

Using ideal gas law:

V_{f} = \frac{n R T}{P} = \frac{0.895 \times 8.314 \times 298}{1 \times 1 0 ^{5}}

V_{f} = 0.02218

m³

Work done by gas =

- W = P_{e x t} V_{f} = 1 \times 1 0^{5} \times 0.02218 = 2218

J

Answer: 2218

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Chemical Thermodynamics

Solution

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