At 25^ C , when 1 mole of butane is heated, CO_2 and liquid H_2 O are formed, then the work done is

Correct answer: B. 85.6 ~L atm aligned & C_4 H_ 10(g) +(13)/(2) O_ 2(g) 4 CO_ 2(g) +5 H_2 O_ (l) & Δ n_g=4-(1+(13)/(2))=-(7)/(2) & w=-Δ n_g R T=-(-(7)/(2) × 0.0821 × 298) & =85.63 L atm aligned

AIIMS 201925 May

Exam: AIIMS
Year: 2019
Date: 25 May
Subject: Chemistry
Chapter: Thermodynamics (C)

AIIMS 2019 · 25 May · Chemistry · Thermodynamics (C)

At 25^ C , when 1 mole of butane is heated, CO_2 and liquid H_2 O...

25^{\circ} \mathrm{C}

, when 1 mole of butane is heated,

\mathrm{CO}_2

and liquid

\mathrm{H}_2 \mathrm{O}

are formed, then the work done is

Options

A $75.6 \mathrm{~L}$ atm
B $85.6 \mathrm{~L}$ atm
C $50.3 \mathrm{~L}$ atm
Dnone of these.

Correct answer

B. $85.6 \mathrm{~L}$ atm

Step-by-step solution

\begin{aligned} & \text{ } \mathrm{C}_4 \mathrm{H}_{10(g)}+\frac{13}{2} \mathrm{O}_{2(g)} \longrightarrow 4 \mathrm{CO}_{2(g)}+5 \mathrm{H}_2 \mathrm{O}_{(l)} \\ & \Delta n_g=4-\left(1+\frac{13}{2}\right)=-\frac{7}{2} \\ & w=-\Delta n_g R T=-\left(-\frac{7}{2} \times 0.0821 \times 298\right) \\ & =85.63 \mathrm{ L} \mathrm{ atm}\end{aligned}

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