Question

Ideal Gas: Please show all work and explain

(a) An ideal gas expands adiabatically from a volume of 2.2 × 10-3 m3 to 3.2 × 10-3 m3. If the initial pressure and temperature were 5 pressure Pa temperature (b) In an isothermal process, an ideal gas expands from a volume of 2.2 10-3 m3 to 3.2 10-3 m3. If the initial pressure and temperature were 5.0 x 105 Pa and 280 K, respectively, what are the final pressure (in Pa) and temperature (in K) of the gas? pressure temperature Pa

Answer 1

In adiabatic process the relation between pressure and volume is p_1v^2_1 = p_2v^2_2 p_1/p_2 = v^2_2/v^2_1 p_2 = p_1(v_1/v_2)^2 = 5 times 10^5(2.2 times 10^-3 /3.2 times 10^-3)^5/3 = 2.67 times 10^5 pa T_1V^2_1 - 1 = T_2V^2_2 - 1 T_2 = T_1(V_1/V_2)^R - 1 = 280k(2.2 times 10^-3/3.2 times 10^-3)^5/3 - 1 = 218.11k in isothermal process PV = nRT p_1v_1 = nRT_1 p_2v_2 = nRT_2 in isothermal process T_1 = T_2 p_1v_1 = p_2v_2 p_2 = p_1v_1/v_2 = 5 times 10^5(2.2 times 10^-3)/3.2 times 10^-3 = 3.43 times 10^5 pa T = 280k