Question

If an ideal gas expands isothermally from 50 L to
100 L against 2 atm external pressure, then values
of W, AU and Q respectively will be
[NCERT Pg. 166)
(1) -100 L atm, zero and +100 L atm
(2) -50 L atm, zero and +50 L atm
(3) +100 L atm, zero and -100 L atm
(4) +50 L atm, zero and -50 L atm.



ANSWER::
w = -p ext v
= -2×(100-50)
= -2×50
= -100L atm.
Gas expands isothermally.
so,
U=0
U= q+ w
0=q-100
___________
➡️ q=100 L atm. ⬅️
___________
​

Answer 1

Answer:

w=-pext.∆v

 =-2(100-50)

 =-100 L atm

∆U=0 because expansion takes place.

∆U=q+w

0=q-100

 q=100 L atm

Answer 2

Answer: The correct answer is option $(1)$.

$w = -100 L atm, U = 0, q = 100 L atm$

Explanation:

When a gas expands isothermally (at constant temperature), work done is calculated as:

$W = -P_{ext}$ ΔV

where $P_{ext}$ is the external pressure and ΔV is the volume expansion.

$W = - 2 atm (100L-50L) = -100 L atm$

There would be change in internal energy as the temperature is constant.

ΔU $= 0$

Now applying first law of thermodynamics to calculate heat involved in this isothermal expansion of an ideal gas:

 $U = q + w$

As $U = 0$

$q = -w = - (-100 L atm) = 100 L atm$.

Therefore correct answer is option $(1)$.

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