Half mole of an ideal gas ( = 5/3 ) is taken through the cycle abcda as shown in the figure. Take R = 25 3 J K ⁻¹ mol ⁻¹ . Determine the temperature of the gas in the states a , b , c and d .

Step-by-step solution

Using the ideal gas equation PV = nRT , the temperature at any state is given by T = PV nR . Given: n = 0.5 mol R = 25 3 J K ⁻¹ mol ⁻¹ nR = 0.5 25 3 = 25 6 J K ⁻¹ For state a : P_a = 100 kPa = 10^5 Pa V_a = 5000 cm ^3 = 5 10⁻³ m ^3 T_a = P_a V_a nR = 10^5 5 10⁻³ 25 6 = 500 6 25 = 120 K For state b : P_b = 100 kPa = 10^5 Pa V_b = 10000 cm ^3 = 10 10⁻³ m ^3 T_b = P_b V_b nR = 10^5 10 10⁻³ 25 6 = 1000 6 25 = 240 K For state c : P_c = 200 kPa = 2 10^5 Pa V_c = 10000 cm ^3 = 10 10⁻³ m ^3 T_c = P_c V_c nR = 2 10^5 10 10⁻