10 mole of oxygen is heated at constant volume from 30^circtextC$30^{\circ}\text{C}$ to 40^circtextC$40^{\circ}\text{C}$. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure, C_p = 7text cal./mol ^circtextC$C_{p} = 7\text{ cal./mol }^{\circ}\text{C}$ and R = 2text cal./mol ^circtextC$R = 2\text{ cal./mol }^{\circ}\text{C}$.)

### Related Formula Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ C_v = C_p - R$C_v = C_p - R$ ### Core Logic Given values: n = 10text moles$n = 10\text{ moles}$ Delta T = 40^circtextC - 30^circtextC = 10^circtextC$\Delta T = 40^{\circ}\text{C} - 30^{\circ}\text{C} = 10^{\circ}\text{C}$ C_p = 7text cal/molcdot^circtextC$C_p = 7\text{ cal/mol}\cdot^{\circ}\text{C}$ R = 2text cal/molcdot^circtextC$R = 2\text{ cal/mol}\cdot^{\circ}\text{C}$ First, find C_v$C_v$ using Mayer's relation: C_v = C_p - R = 7 - 2 = 5text cal/molcdot^circtextC$$C_v = C_p - R = 7 - 2 = 5\text{ cal/mol}\cdot^{\circ}\text{C}$$ ### Step 1: Calculate Internal Energy Change Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ Delta U = 10 times (7 - 2) times (40 - 30)$$\Delta U = 10 \times (7 - 2) \times (40 - 30)$$ Delta U = 10 times 5 times 10 = 500text cal$$\Delta U = 10 \times 5 \times 10 = 500\text{ cal}$$ ### Pattern Recognition Change in internal energy of an ideal gas is ALWAYS Delta U = n C_v Delta T$\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use C_v = C_p - R$C_v = C_p - R$ when C_p$C_p$ is given. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases