Question

✡️ An ideal gas having initial pressure P, Volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2.k

( i ). How many degrees of freedom do the gas molecules have?

( ii ). Obtain the work done by the gas during expansion as a function of the initial pressure P and volume V.

Answer 1

The adiabatic equation of a perfect gas is given by equation:

TV∧(γ-1)=constant

orT1V1∧(γ-1)=T1V2∧(γ-1)

hereT1=T1V1=V1V2=5.66V and

T2=(T/2)

TV∧(γ-1)=[T/2] (5.66V)∧(γ-1)

2=(5.66)∧(γ-1)

Taking log on both sides, we get :

(γ-1)log5.66 =log 2(γ-1)

=log 2/log 5.66
=1+0.3010/0.7528
γ=1+0.4=1.4



B) The work done during adiabatic process,

W=[1/(γ-1)] x [P1V1-P2V2]from perfect gas relation ,

P1V1/T1=P2V2/T2

or P2=(T2/T1)(P1V1)/T1

P2=(T/2)/5.66V[PV/T]

P2=P/5.66X2=P/11.32

w=(1/1.4-1)x [ PV-P/11.32X5.66V]

=PV/0.4(1-0.5)

=0.5/0.4 PV
W=5/4PV
=1.25PV

Answer 2

ᴛʜᴇ ᴀᴅɪᴀʙᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴀ ᴘᴇʀғᴇᴄᴛ ɢᴀs ɪs ɢɪᴠᴇɴ ʙʏ ᴇǫᴜᴀᴛɪᴏɴ:

ᴛᴠ∧(γ-1)=ᴄᴏɴsᴛᴀɴᴛ

ᴏʀᴛ1ᴠ1∧(γ-1)=ᴛ1ᴠ2∧(γ-1)

ʜᴇʀᴇᴛ1=ᴛ1ᴠ1=ᴠ1ᴠ2=5.66ᴠ ᴀɴᴅ

ᴛ2=(ᴛ/2)

ᴛᴠ∧(γ-1)=[ᴛ/2] (5.66ᴠ)∧(γ-1)

2=(5.66)∧(γ-1)

ᴛᴀᴋɪɴɢ ʟᴏɢ ᴏɴ ʙᴏᴛʜ sɪᴅᴇs, ᴡᴇ ɢᴇᴛ :

(γ-1)ʟᴏɢ5.66 =ʟᴏɢ 2(γ-1)

=ʟᴏɢ 2/ʟᴏɢ 5.66
=1+0.3010/0.7528
γ=1+0.4=1.4

ʙ) ᴛʜᴇ ᴡᴏʀᴋ ᴅᴏɴᴇ ᴅᴜʀɪɴɢ ᴀᴅɪᴀʙᴀᴛɪᴄ ᴘʀᴏᴄᴇss,

ᴡ=[1/(γ-1)] x [ᴘ1ᴠ1-ᴘ2ᴠ2]ғʀᴏᴍ ᴘᴇʀғᴇᴄᴛ ɢᴀs ʀᴇʟᴀᴛɪᴏɴ ,

ᴘ1ᴠ1/ᴛ1=ᴘ2ᴠ2/ᴛ2

ᴏʀ ᴘ2=(ᴛ2/ᴛ1)(ᴘ1ᴠ1)/ᴛ1

ᴘ2=(ᴛ/2)/5.66ᴠ[ᴘᴠ/ᴛ]

ᴘ2=ᴘ/5.66x2=ᴘ/11.32

ᴡ=(1/1.4-1)x [ ᴘᴠ-ᴘ/11.32x5.66ᴠ]

=ᴘᴠ/0.4(1-0.5)

=0.5/0.4 ᴘᴠ
ᴡ=5/4ᴘᴠ
=1.25ᴘᴠ