Question

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$\huge\underline{Question}$
1 mole of gas occupying 5L volume is expanded against a constant external pressure of 1atm to a volume of 20 L. The work done by a system is equal to
Options:
(a) -1.519 x 10²J
(b)-1.519 x 10³ J
(c) +1.519 x 10³J
(d) -1.591 ×10³ J​

Answer 1

[b] - 1.519 × 10³ J .

$\large\mathcal\green{\underbrace{\orange{SOLUTION:-}}}$

GIVEN :-

• 1 mole of gas occupying 5L volume is expanded against a constant external pressure of 1 atm to a volume of 20L .

✍️ Where,

• $\rm{\red{P_{ext}}}$ = 1 atm .

• $\rm{\red{V_1}}$ = 5 L .

• $\rm{\red{V_2}}$ = 20 L .

TO FIND :-

• The work done by the system .

CALCULATION :-

✍️ We have know that,

As work is done against constant external pressure, the process is Irreversible .

$\red\bigstar\:\mathcal{\purple{\boxed{\blue{Work\:done\:=\:-\:P_{ext}\:.\:\triangle{V}\:}}}}$

Where,

• ∆V = Change in Volume .

• $\rm{\red{\implies}\:V_2\:-\:V_1\:=\:20\:-\:5\:}$

• $\rm{\red{\implies}\:\triangle{V}\:=\:15\:L}$

$\rm{\implies\:W\:=\:-\:1\:\times{15}\:}$

$\rm{\pink{\implies\:W\:=\:-\:15\:atm.L\:}}$

✍️ Conversion into joule unit .

• [1 atm . L = 101.3 J]

$\rm{\implies\:W\:=\:-\:15\:\times{101.3}\:}$

$\rm{\implies\:W\:=\:-\:1519.5\:J\:}$

$\red\checkmark\:\rm{\purple{\boxed{Work\:done\:=\:-\:1.519\times{10^3}\:J\:}}}$--(b)

Answer 2

Answer:

ᴘ ᴇxᴛ=1 ᴀᴛᴍ

V 1=5L

⇒V2 =20 L

As ᴡᴏʀᴋ ɪs ᴅᴏɴᴇ ᴀɢᴀɪɴsᴛ ᴄᴏɴsᴛᴀɴᴛ ᴇxᴛᴇʀɴᴀʟ ᴘʀᴇssᴜʀᴇ, ᴛʜᴇ ᴘʀᴏᴄᴇss ɪs ɪʀʀᴇᴠᴇʀsɪʙʟᴇ.

W=−ᴘ ᴇxᴛ ΔV

=−1 ᴀᴛᴍ[20-5]L=-15 ᴀᴛᴍ L

=-15 ×101.3 J[1 L ᴀᴛᴍ=101.3 J]

=−1.519 J=−1.519×10 ^3

J.

Hᴇɴᴄᴇ, ᴛʜᴇ ᴄᴏʀʀᴇᴄᴛ ᴏᴘᴛɪᴏɴ ɪs B

Explanation:

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