What is the change in internal energy when a gas contracts from $325 \mathrm{~ml}$ to $125 \mathrm{~ml}$ at a constant pressure of $2 \mathrm{~bar},$ while at the same time being cooled by removing $124 \mathrm{~J}$ heat?

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

What is the change in internal energy when a gas contracts from

325 \mathrm{~ml}

125 \mathrm{~ml}

at a constant pressure of

2 \mathrm{~bar},

while at the same time being cooled by removing

124 \mathrm{~J}

heat?

View answer and explanation

Correct option: B

-84 \mathrm{~J}

Option B (-84 J) is correct because it is calculated using the first law of thermodynamics, considering both the heat removed and the work done due to contraction.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

The internal energy of one mole of a gas is: An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If

{T_i}

is the initial temperature and

{T_f}

is the final temperature, which of the following statements is correct? For which of the process, ∆S is negative? The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing pressure, P is variable, is given by: Assertion :- At constant pressure of the change

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

work done is negative.

Reason :- During phase transition work done is always negative. For the reaction,

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

300 \mathrm{~K}

. Hence,

\Delta G