Grade 11Thermal Physics

Prove that work done by an ideal gas in an adiabatic process is W =[P(initial)V(initial)- P(final)V(final)]/(gamma-1), using integration [integral of PdV]

Arnab Mandal

14 Years agoGrade 11

1 Answer

Aman Bansal

14 Years ago

Dear Arnab,

In an adiabatic quasi-static process we can write the formula

PV^Y = constant

constant = K for simplification

Since its adiabatic no heat change so Q=0

Using the first law of thermo

Q= ΔU -W

We know that W = -PdV
and P= K/V^Y

so...
W = ΔU
W = -PdV
W = -(K/V^Y)*dV
W = -K∫(1/V^Y)*dV
W = -K[V^(1-Y)/(1-Y)]*∫dV
W = -(K/(1-Y))[Vf^(1-Y) - Vi^(1-Y)]
W = -(K/(1-Y))[Vf^(-Y)*Vf - Vi^(-Y)*Vi]
W = -(1/(1-Y))[((Vf*K)/(Vf^Y)) - ((Vi*K)/(Vi^Y))]

since Pi = K/Vi^Y and Pf = K/Vf^Y sub those in

W = -(1/(1-Y))(Vf*Pf - Vi*Pi)
Times this by (-1/-1)

and we get

W = (PfVf - PiVi)/(Y-1)

This is where I get to not sure where to go from here to make this into

W= (PfVf)/(Y-1)[1 - (Pi/Pf)^((Y-1)/Y)]

Best Of Luck...!!!!

Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

A container filled with a sample of an Idea gas at the pressure of 1.5atm. the gas is compressed isothermal ly to one fourth of its original volume. What is the new pressure of the gas?

thermal physics

1 Answer Available

Last Activity: 3 Years ago

how do we know the electric field due to an electric dipole of dipole moment P at distance r from the centre of dipole is fiven by .. where theta is angle between p amd r vector..

thermal physics

1 Answer Available

Last Activity: 4 Years ago

A A mole of helium gas initially at stp undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by gas is

thermal physics

1 Answer Available

Last Activity: 4 Years ago

5000 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 500 K, while the gas performs 7500 J of work. What is the final temperature of the gas? step by step please

thermal physics

1 Answer Available

Last Activity: 4 Years ago

Please help me with the answer of this question related to radiation graph

thermal physics

1 Answer Available

Last Activity: 4 Years ago

View all Questions

Prove that work done by an ideal gas in an adiabatic process is W =[P(initial)*V(initial)- P(final)*V(final)]/(gamma-1), using integration [integral of PdV]

Arnab Mandal

1 Answer

Aman Bansal

Other Related Questions on thermal physics

A container filled with a sample of an Idea gas at the pressure of 1.5atm. the gas is compressed isothermal ly to one fourth of its original volume. What is the new pressure of the gas?

how do we know the electric field due to an electric dipole of dipole moment P at distance r from the centre of dipole is fiven by .. where theta is angle between p amd r vector..

A A mole of helium gas initially at stp undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by gas is

5000 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 500 K, while the gas performs 7500 J of work. What is the final temperature of the gas? step by step please

Please help me with the answer of this question related to radiation graph

Prove that work done by an ideal gas in an adiabatic process is W =[P(initial)V(initial)- P(final)V(final)]/(gamma-1), using integration [integral of PdV]