Question

When a sample of argon (for which gamma=5/3) at 100KPA expands reversibly and adiabatically to twice its initial volume the final pressure will be:

Answer 1

For reversible adiabatic expansion

We have

$$PV^{\gamma}=constant$$

So

$$P_2=P_1(\frac{V_1}{V_2})^\gamma=100\times(\frac{V_1}{2V_1})^{1.66}\\=100\times0.5^{1.66}\\100\times0.31644=31.644\ kPa$$