A gas expands against a constant external pressure of $2.00 a t m$ , increasing its volume by $3.40 L$ . Simultaneously, the system absorbs $400 J$ of heat from its surroundings. What is $Δ E$ , in joules, for this gas?

- 689

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- 289

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+ 400

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+ 289

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Solution

The correct option is B $- 289$
Solution:- (B) $- 289 J$
As the heat is absorbed by the system, therefore,
$q = + 400 J$
Now as we know thta,
$W = - P_{e x t .} Δ V$
Given:-
$P_{e x t .} = 2 a t m$
$Δ V = 3.4 L$
$∴ W = - 2 \times 3.4 = - 6.8 L - a t m = - 689 J$
Now, from first law of thermodynamics,
$Δ E = q + W$
$\Rightarrow Δ E = 400 + (- 689) = - 289 J$
Hence the value of $Δ E$ for the gas is $- 289 J$ .

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A gas expands against a constant external pressure of 2.00 atm, increasing its volume by 3.40 L. Simultaneously, the system absorbs 400 J of heat from its surroundings. What is ΔE, in joules, for this gas?

A gas expands against a constant external pressure of $2.00 a t m$ , increasing its volume by $3.40 L$ . Simultaneously, the system absorbs $400 J$ of heat from its surroundings. What is $Δ E$ , in joules, for this gas?