Three moles of an ideal gas at 300 K are isothermally expanded to five times its volume and heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 83.14 KJ heat is required. Calculate the ratio C_p//C_v of the gas. log. 5 = 1.61 and R = 8.3 Joules/mole K.

Three moles of an ideal gas at 300 K are isothermally expanded to five

1.	Three moles of an ideal gas at 300 K are isothermally expanded to five times its volume and heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 83.14 KJ heat is required. Calculate the ratio C_p//C_v of the gas. log. 5 = 1.61 and R = 8.3 Joules/mole K.
Answer» Solution :For the isothermal expansion, by Boyle.s law, we have `P_1 V_1=P_2 V_2or(P_1)/(P_2) = (V_2)/(v_1) = (5V_1)/( V_1) = 5` Now, applying gas law at constant volume, we GET `(P_2)/(T_2)=(P_3)/(T_3)( thereforeP_3 = P_1)(or)(P_2)/(T_2) =(T_2)/(T_3) = (T_1)/(T_3)` (As FIRST expansion was isothermal, HENCE `T_1 =T_2`.) `threfore(T_3)/(T_1) =(P_1)/(P_2) = 5or T_3 = 5T_1 = 5 xx300= 1500K` Now ` dQ= dU+ DW= nC_vd Tint_( V_i)^(v_2) PdV ` But `P= nRT_1 //V`, where `mu`is the number of moles of the gas. ` thereforedQ= nC _v dT + nRT_1log _e(V_2 //V_1)` Now `C_p-C_v=Rand(C_p -C_v) = gamma. so C_v = R // ( gamma -1)` `therefored Q=( nR)/( gamma -1) dT+ nRT_1log_e(V_2)/(v_1)` SUBSTITUTING the values : `83.14 xx 10^3` Joule `=( 3 xx 8.3 )/( gamma -1)` `(1500- 300 ) +8.3300log_e5 (or)83140 =( 29880)/( gamma -1) + 12027( log_e 5=1.61 ]thereforegamma= 1.42`