Question

# A gas is allowed to expand in a well insulated container against constant external pressure of $2.5\mathrm{atm}$ from an initial volume of $2.50\mathrm{L}$ to a final volume of $4.50\mathrm{L}$. The change in internal energy $\mathrm{\Delta }\mathrm{U}$ of the gas in joules will be ?

A

$+505\mathrm{J}$

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B

$+1136.25\mathrm{J}$

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C

$-500\mathrm{J}$

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D

$-505\mathrm{J}$

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Solution

## The correct option is D $-505\mathrm{J}$Explanation for the correct option:Option(D):$-505\mathrm{J}$Step 1: Analyzing Given data:The constant external pressure is given as= $2.5\mathrm{atm}$The initial volume =$2.50\mathrm{L}$ and final volume = $4.50\mathrm{L}$Step 2: Formula for calculating $\mathrm{\Delta U}$:According to the first law of thermodynamics, the formula related to the change in the internal energy can be given as:$\mathrm{\Delta U}=\mathrm{q}+\mathrm{W}$(where$\mathrm{\Delta U}$ = change in internal energy of the system$\mathrm{q}=$ heat transfer between the system$\mathrm{W}=$work done by the system)Step 3: Calculation of $\mathrm{\Delta U}$:Also, as per the given question, the gas is allowed to expand in a well-insulated container, hence, heat transfer i.e $\mathrm{q}=0$Hence,$\mathrm{\Delta U}=\mathrm{W}=-\mathrm{P\Delta V}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta U}=\mathrm{W}=-2.5\mathrm{atm}\mathrm{x}\left(4.50\mathrm{L}–2.50\mathrm{L}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta U}=\mathrm{W}=-5.0\mathrm{L}\mathrm{atm}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta U}=\mathrm{W}=-5.0\mathrm{L}\mathrm{atm}\mathrm{x}\left(101.33\mathrm{J}/1\mathrm{L}\mathrm{atm}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta U}=-505\mathrm{J}$Hence, the change in internal energy $\mathrm{\Delta }\mathrm{U}$ of the gas in joules will be $-505\mathrm{J}$.Thus, option D is correct.

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