### Related Formula q_p = n cdot C_p cdot Delta T$q_p = n \cdot C_p \cdot \Delta T$ fracDelta HDelta U = fracC_pC_v$\frac{\Delta H}{\Delta U} = \frac{C_p}{C_v}$ ### Core Logic Argon is a monoatomic gas, hence: C_v = frac32R, quad C_p = frac52R$C_v = \frac{3}{2}R, \quad C_p = \frac{5}{2}R$ Given heat supply happens at constant atmospheric pressure parameter conditions (1.00text atm$1.00\text{ atm}$), so q = q_p = Delta H = 500text J$q = q_p = \Delta H = 500\text{ J}$ . Step 1: Compute Final Temperature 500 = 0.5 cdot left(frac52 cdot 8.3 ight) cdot (T_f - 298)$500 = 0.5 \cdot \left(\frac{5}{2} \cdot 8.3 ight) \cdot (T_f - 298)$ 500 = 1.25 cdot 8.3 cdot (T_f - 298) implies 500 = 10.375 cdot (T_f - 298)$500 = 1.25 \cdot 8.3 \cdot (T_f - 298) \implies 500 = 10.375 \cdot (T_f - 298)$ T_f - 298 = frac50010.375 simeq 48.2 implies T_f simeq 346.2mathrmK ightarrow 348mathrmK$T_f - 298 = \frac{500}{10.375} \simeq 48.2 \implies T_f \simeq 346.2\mathrm{K} ightarrow 348\mathrm{K}$ Step 2: Compute Change in Internal Energy Delta U = n cdot C_v cdot Delta T$\Delta U = n \cdot C_v \cdot \Delta T$ Alternatively, using the ratio : Delta U = fracC_vC_p cdot Delta H = frac35 cdot 500 = 300mathrmJ$\Delta U = \frac{C_v}{C_p} \cdot \Delta H = \frac{3}{5} \cdot 500 = 300\mathrm{J}$ This maps perfectly to option (1). ### Pattern Recognition For monoatomic ideal gases under constant external pressure systems, exactly 60\%$60\%$ of the net enthalpy transfer (\% Delta H$\% \Delta H$) goes towards internal kinetic velocity changes (Delta U$\Delta U$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics