MHT CET · Chemistry · Thermodynamics (C)

What is the amount of work done when 0.5 mole of methane, $C H_{4} (g)$ is subjected to combustion at 300 K? (Given, $R = 8.314 J K^{- 1} m o l^{- 1}$ )

A - 2494 J
B - 4988 J
C + 4988 J
D +2494 J

Verified Solution

Answer & Solution

Correct Answer

(D) +2494 J

Step-by-step Solution

Detailed explanation

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

Amount of work done

= - ∆ n R T

1 mole of methane reacts with 2 moles of oxygen to produce 2 moles of

C O_{2}

Hence, 0.5 mole of

C H_{4}

reacts with 1 mole of

O_{2}

to produce 1 mole of

C O_{2}

∴ n_{1} =

moles of reactants = 1.5

n_{2} =

moles of product = 1

∴ ∆ n = n_{2} - n_{1} = 1 - 1.5 = - 0.5

w = - (- 0.5) \times 300 \times 8.314 J K^{- 1} m o l^{- 1}

= + 2494 J

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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What is the amount of work done when 0.5 mole of methane, CH4g is subjected to combustion at 300 K? (Given, R=8.314 J K-1 mol-1 )

Step-by-step Solution

What is the amount of work done when 0.5 mole of methane, $C H_{4} (g)$ is subjected to combustion at 300 K? (Given, $R = 8.314 J K^{- 1} m o l^{- 1}$ )