Answer to Question #86128 in Molecular Physics | Thermodynamics for mmmmmmmmmmmm

Question

A monatomic ideal gas that is initially at a pressure of 1.5x105 N/m2 and has a volume of
0.08 m3 is compressed adiabatically to a volume of 0.04 m3 (a) What is the final pressure?
(b) How much work is done by the gas? (c) What is the ratio of the final temperature of
the gas to its initial temperature?

Answer 1

(a) The equation of the reversible adiabatic process

$$P_1V_1^{\gamma}=P_2V_2^{\gamma}, \quad \gamma=\frac{C_P}{C_V}=\frac{5}{3}$$

So, the final pressure

$$P_2=P_1\left(\frac{V_1}{V_2}\right)^{5/3}=1.5\times 10^5\left(\frac{0.08}{0.04}\right)^{5/3}=4.76\times 10^5\;\rm{N/m^2}$$

(b) The work done

$$W=-\Delta U=-\frac{3}{2}(P_2V_2-P_1V_1)=-\frac{3}{2}(P_2V_2-P_1V_1)$$

$$=-\frac{3}{2}(4.76\times 10^5\times 0.04-1.05\times 10^5\times 0.08)=-10640\;\rm{J}$$

(c)

$$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}=\left(\frac{0.08}{0.04}\right)^{5/3-1}=1.59$$