Work done by gas in adiabatic process

[collectedsoul]

In an adiabatic process PV^γ=constant

Now I thought the work done by an ideal gas in an adiabatic process was given by the equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

But while doing a GRE question the answer was given as P_fV_f - P_iV_i / 1 - γ

Is this correct (it must be!)? And if so why haven't I seen this equation anywhere before?

 

[CAF123]

I think the two are equivalent. Reexpress the formula in the link using, for an adiabatic process, $$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$

 

[Andrew Mason]

In an adiabatic process PV^γ=constant

Now I thought the work done by an ideal gas in an adiabatic process was given by the equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

But while doing a GRE question the answer was given as P_fV_f - P_iV_i / 1 - γ

Is this correct (it must be!)? And if so why haven't I seen this equation anywhere before?

Did you mean that the GRE answer was W = (P_fV_f - P_iV_i) / (1 - γ) ?

Since it is a reversible adiabatic process (Q=0), the work done BY the gas has to be equal and opposite to its change in internal energy: W = - ΔU (first law).

For an ideal gas, ΔU = nCvΔT = nCv(γ-1)/(γ-1)ΔT = nRΔT/(γ-1) = Δ(PV)/(γ-1) = (PfVf - PiVi)/(γ-1).

AM

 

[CAF123]

Alternatively, $$P_i V_i^{\gamma} = P_f V_f^{\gamma} = K \Rightarrow KV_i^{1-\gamma} = P_iV_i\,\,\text{and}\,\,KV_f^{1-\gamma} = P_fV_f$$
Then sub in.