Q.An ideal gas of volume 1.0 L is adiabatically compressed to (1/15)ᵗʰ of its initial volume. Its initial pressure and temperature are 1.01 × 10⁵ Pa and 27 °C respectively. Given Cᵥ for ideal gas = 20.8 J/mol·K and γ = 1.4. Calculate

Adiabatic: PVγ = const → Pf = Pi(Vi/Vf)^γ; W = (PfVf − PiVi)/(1−γ); Tf = Ti(Vi/Vf)^(γ−1). …

1. Final pressure. For an adiabatic change PVγ = constant, so

   $P_f = P_i\left(\frac{V_i}{V_f}\right)^{\gamma} = (1.01\times10^{5})(15)^{1.4} = 44.8\times10^{5}\ \mathrm{Pa}\ (\approx 45\ \mathrm{atm})$

2. Work done. With Vf = Vi/15 = 1/15 L,

   $W = \frac{P_fV_f - P_iV_i}{1-\gamma} = -494\ \mathrm{J}$

   The work is negative — work is done ON the gas during compression.
3. Final temperature.

   $T_f = T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1} = (300)(15)^{0.40} = 886\ \mathrm{K} = 613\ ^\circ\mathrm{C}$

   The gas heats up strongly because no heat escapes — the same principle as the ignition of fuel in a diesel engine. …

Use PVγ = const for Pf and Tf; W = (PfVf−PiVi)/(1−γ); contrast with the isothermal resu …

Using γ in place of (γ−1) for the temperature relation, or expecting the adiabati …