Three moles of an ideal gas () and 2 moles of another ideal gas () are taken in a vessel and compres — Thermodynamics and Thermochemistry Chemistry Question
Question
Three moles of an ideal gas ($C_{P,m} = 2.5R$) and 2 moles of another ideal gas ($C_{P,m} = 3.5R$) are taken in a vessel and compressed reversibly and adiabatically. In this process, the temperature of gaseous mixture increased from 300 K to 400 K. The increase in internal energy of gaseous mixture (in cal) is
Answer: 1900
💡 Solution & Explanation
For gas 1, $C_{V,m} = 1.5R$. For gas 2, $C_{V,m} = 2.5R$. Total heat capacity $C_V = n_1 C_{V,1} + n_2 C_{V,2} = 3(1.5R) + 2(2.5R) = 4.5R + 5.0R = 9.5R$. $\Delta U = C_V \Delta T = 9.5R \times (400 - 300) = 950R$. Using $R = 2\text{ cal/K-mol}$, $\Delta U = 950 \times 2 = 1900\text{ cal}$.
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