Question

1 mole of a mono-atomic gas (y=5/3) at 27°c is adiabatically compressed in a reversible process from an initial pressure 1atm to a final pressure of 50atm. Calculate the resulting difference of temperature

Answer 1

The equation of the adiabatic process

PV^γ=const

Using ideal gas equation of state,

PV=nRT

we obtain

P^(1-γ) T^γ=const

Therefore

〖P_i〗^(1-γ) 〖T_i〗^γ=〖P_f〗^(1-γ) 〖T_f〗^γ

T_f=T_i (P_i/P_f )^((1-γ)/γ)

=(27℃+273) (1/50)^((1-5/3)/(5/3))=1435 K=1162 ℃

So, the resulting difference of temperature

Δt=1162℃-27℃=1135℃