Thermodynamics

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New answer posted

5 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

 C (s)+O2 (g)CO2 (g)Δng=0

qC=20*2=40kJ

Number of moles of C = 2.412=0.2moles

ΔHC=40kJ0.2mole=200kJ/mole

New answer posted

5 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

ΔH=ΔU+ΔngRT

ΔH=59.6+1*8.314*300*103=57.10kJ/mol

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

              44.4 litres of He 4 4 . 8 2 2 . 4 m o l e s

            η = 2    moles of He

              Q =   η C v Δ T (for fixed capacity (v = constant)

              =   2 * R γ 1 Δ T

              =   2 * R 5 3 1 * 2 0

= = 2 * 3 R 2 * 2 0

 Q = 60R = 60 * 8.3

  Q = 498 J

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Cv=n*R2

Cp=Cv+R= (n+2)R2

CvCP=n (n+2)

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

 ΔQ=ΔU+ΔW

Q=ΔU+Q5ΔU=4Q5=nCvΔT4Q5=5R2ΔTΔT=8Q25R

Q=ncΔT=1*C*8025RC=25R8x=25

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Constant Entropy means Adiabatic process

Pvy=C

P1v1y=P2v2y

P2=P1 (v1v2)y=P (v18v)y

P2=P (8)53

P2 = 32 P

New answer posted

5 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

V=kT23

TV32=K

γ1=3/2

γ=1/2

Work done = nRΔTy+1=1*R*903/2=6R

n = 60

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 Δu=nCVΔT

=n3R2ΔT

=7*32*8.3*40

=3486J

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

U = 3PV + 4

nCvT=3PV+4

n*fRT2=3PV+4

f=6+8PV

f>6polyatomic

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Mono atomic ? Cv=3R2Cp=5R2

Di- atomic ? CV=5R2Cp=7R2

(Rigid)

Di-atomic ? Cv=7R2Cp=9R2

(Non-Rigid)

Tri-atomic ? CV=3RCP=4R

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