One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27^circ$27^{\circ}$ C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27^circ$27^{\circ}$ C doing the same amount of work W, then its final temperature will be (close to) ____ ^circ${}^{\circ}$ 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

### Related Formula W = PDelta V = nRDelta T$$W = P\Delta V = nR\Delta T$$ Q = nC_pDelta T$$Q = nC_p\Delta T$$ C_p = left(fracf2 + 1right)R$$C_p = \left(\frac{f}{2} + 1\right)R$$ ### Core Logic For an isobaric (constant pressure) process, work done is given by: W = nRDelta T = 100 text J$$W = nR\Delta T = 100 \text{ J}$$ For a diatomic gas, the degrees of freedom f = 5$f = 5$. Thus, the molar heat capacity at constant pressure is: C_p = left(frac52 + 1right)R = frac72R$$C_p = \left(\frac{5}{2} + 1\right)R = \frac{7}{2}R$$ ### Step 1: Calculating Heat Transfer The heat supplied to the gas is: Q = nC_pDelta T = n left(frac72Rright)Delta T = frac72 (nRDelta T)$$Q = nC_p\Delta T = n \left(\frac{7}{2}R\right)\Delta T = \frac{7}{2} (nR\Delta T)$$ ### Step 2: Final Conclusion Substitute the value of work done: Q = frac72 times (100) = 350 text J$$Q = \frac{7}{2} \times (100) = 350 \text{ J}$$ ### Pattern Recognition In an isobaric process, the ratio of Work : Internal Energy Change : Heat Added (W : Delta U : Q$W : \Delta U : Q$) is always 2 : f : (f+2)$2 : f : (f+2)$. For diatomic gases, f=5$f=5$, so the ratio is 2:5:7$2:5:7$. Thus Q = frac72W$Q = \frac{7}{2}W$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

### Related Formula TV^gamma-1 = textconstant$$TV^{\gamma-1} = \text{constant}$$ ### Core Logic

Using adiabatic relation TV^gamma-1 = textconstant$TV^{\gamma-1} = \text{constant}$: T_1 V_1^gamma-1 = T_2 V_2^gamma-1$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$ T(V)^gamma-1 = left(fracT4right)(8V)^gamma-1$$T(V)^{\gamma-1} = \left(\frac{T}{4}\right)(8V)^{\gamma-1}$$ 4 = 8^gamma-1 implies 2^2 = 2^3(gamma-1) implies gamma = frac53$$4 = 8^{\gamma-1} \implies 2^2 = 2^{3(\gamma-1)} \implies \gamma = \frac{5}{3}$$ Since gamma = 5/3$\gamma = 5/3$, the gas is monoatomic (Helium / He). ### Pattern Recognition Sees: Adiabatic expansion with volume and temperature changes. Shortcut: Apply TV^gamma-1$TV^{\gamma-1}$ relation to determine adiabatic exponent gamma$\gamma$. Check: Matches option (4). ✓ ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula fracP_1 V_1T_1 = fracP_2 V_2T_2$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ T(mathrmK) = T(^circmathrmC) + 273$$T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273$$ ### Core Logic Converting initial and final temperatures to Kelvin: T_1 = 27^circmathrmC + 273 = 300 mathrm~K$$T_1 = 27^\circ\mathrm{C} + 273 = 300 \mathrm{~K}$$ T_2 = 77^circmathrmC + 273 = 350 mathrm~K$$T_2 = 77^\circ\mathrm{C} + 273 = 350 \mathrm{~K}$$ Applying ideal gas law relation: frac2 times 60300 = fracP_2 times 20350$$\frac{2 \times 60}{300} = \frac{P_2 \times 20}{350}$$ frac120300 = frac20 P_2350$$\frac{120}{300} = \frac{20 P_2}{350}$$ frac25 = frac20 P_2350 implies 20 P_2 = 140 implies P_2 = 7 mathrm~atm$$\frac{2}{5} = \frac{20 P_2}{350} \implies 20 P_2 = 140 \implies P_2 = 7 \mathrm{~atm}$$ ### Step 1: Final Conclusion The final pressure is 7$7$ atmospheric pressure. ### Pattern Recognition Combined Gas Law: P_2 = P_1 left(fracV_1V_2right) left(fracT_2T_1right)$P_2 = P_1 \left(\frac{V_1}{V_2}\right) \left(\frac{T_2}{T_1}\right)$. P_2 = 2 times left(frac6020right) times left(frac350300right) = 2 times 3 times frac76 = 7text atm$P_2 = 2 \times \left(\frac{60}{20}\right) \times \left(\frac{350}{300}\right) = 2 \times 3 \times \frac{7}{6} = 7\text{ atm}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula Delta Q = Delta U + W$$\Delta Q = \Delta U + W$$ Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ W = P Delta V = P (A cdot Delta x)$$W = P \Delta V = P (A \cdot \Delta x)$$ ### Core Logic For a monoatomic gas like Helium, C_v = frac32R$C_v = \frac{3}{2}R$. Given internal energy is 3nRT$3nRT$, but standard change is Delta U = n C_v Delta T$\Delta U = n C_v \Delta T$. Wait, the problem states internal energy U = 3nRT$U = 3nRT$ (which implies C_v = 3R$C_v = 3R$ for this specific state/setup, or it's a typo in the paper and they mean U = frac32nRT$U = \frac{3}{2}nRT$). Let's follow the solution's lead: Delta U = 3nRDelta T$\Delta U = 3nR\Delta T$ is directly given. ### Step 1: Calculate Change in Internal Energy Using the given relation: Delta U = 3nR Delta T$$\Delta U = 3nR \Delta T$$ Substitute n=1$n=1$, R = frac253 \, mathrmJ/mol cdot K$R = \frac{25}{3} \, \mathrm{J/mol \cdot K}$, and Delta T = 4^circmathrmC$\Delta T = 4^{\circ}\mathrm{C}$: Delta U = 3 times 1 times frac253 times 4 = 100 \, mathrmJ$$\Delta U = 3 \times 1 \times \frac{25}{3} \times 4 = 100 \, \mathrm{J}$$ ### Step 2: Apply First Law of Thermodynamics We know Delta Q = 126 \, mathrmJ$\Delta Q = 126 \, \mathrm{J}$. W = Delta Q - Delta U$$W = \Delta Q - \Delta U$$ W = 126 - 100 = 26 \, mathrmJ$$W = 126 - 100 = 26 \, \mathrm{J}$$ ### Step 3: Calculate Piston Displacement Work done is isobaric since the piston is freely moving: W = P Delta V = P cdot A cdot Delta x$$W = P \Delta V = P \cdot A \cdot \Delta x$$ 26 = 10^5 times 17 times 10^-4 times Delta x$$26 = 10^{5} \times 17 \times 10^{-4} \times \Delta x$$ 26 = 170 cdot Delta x$$26 = 170 \cdot \Delta x$$ Delta x = frac26170 \, mathrmm = frac26170 times 100 \, mathrmcm = 15.3 \, mathrmcm$$\Delta x = \frac{26}{170} \, \mathrm{m} = \frac{26}{170} \times 100 \, \mathrm{cm} = 15.3 \, \mathrm{cm}$$ Closest given option is 15.5 \, mathrmcm$15.5 \, \mathrm{cm}$. ### Pattern Recognition When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified U$U$ definitions given in the prompt, replacing standard C_v$C_v$ logic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics