Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 498: 26

(a) $1.14atm$ (b) $321K$

(a) We can find the final pressure as follows: $p_2=p_1(\frac{V_1}{V_2})^{\gamma}$ We plug in the known values to obtain: $p_2=3(0.5)^{1.40}$ $\implies p_2=1.14atm$ (b) We can find the final temperature as follows: $T_2=(\frac{V_1}{V_2})^{\gamma-1}T_1$ We plug in the known values to obtain: $T_2=(0.5)^{1.40-1}(423)$ $\implies T_2=321K$