An insulated cylinder of volume 60 mathrm~cm^3$60 \mathrm{~cm}^3$ is filled with a gas at 27^circmathrmC$27^\circ\mathrm{C}$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 mathrm~cm^3$20 \mathrm{~cm}^3$ while allowing the temperature to rise to 77^circmathrmC$77^\circ\mathrm{C}$. The final pressure is ____ atmospheric pressure.

### 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 1. First Law of Thermodynamics: dQ = dU + dW$dQ = dU + dW$ In an adiabatic process: dQ = 0 implies dW = -dU$$dQ = 0 \implies dW = -dU$$ 2. Change in Internal Energy: dU = n C_v dT = n C_v (T_2 - T_1)$$dU = n C_v dT = n C_v (T_2 - T_1)$$ ### Core Logic Let's analyze each statement: - **(A) Internal energy is constant:** Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy (U propto T$U \propto T$) must change. - **(B) Work done is equal to the change in internal energy:** Correct in magnitude (|dW| = |dU|$|dW| = |dU|$). By definition, dW = -dU$dW = -dU$, which correlates the magnitude of work to the change in internal energy. - **(C) Product of temperature and volume is constant:** Incorrect. The adiabatic equation of state is T V^gamma-1 = textconstant$T V^{\gamma-1} = \text{constant}$. - **(D) Product of pressure and volume is constant:** Incorrect. The relation is P V^gamma = textconstant$P V^\gamma = \text{constant}$. - **(E) Work done is proportional to (T_2 - T_1)$(T_2 - T_1)$:** Correct. Since dW = -dU = -n C_v (T_2 - T_1)$dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change (T_2 - T_1)$(T_2 - T_1)$. ### Step 1: Determine the correct option Since only statements (B) and (E) are correct, the correct option is (3). ### Pattern Recognition Sees: Characteristics of adiabatic thermodynamic process. Trap: Confusing adiabatic state relations (PV^gamma = C$PV^\gamma = C$, TV^gamma-1 = C$TV^{\gamma-1} = C$) with isothermal state relations (PV = C$PV = C$, T = C$T = C$). Shortcut: First law of thermodynamics under dQ=0$dQ=0$ strictly enforces |dW| = |dU|$|dW| = |dU|$, which validates statement B and E immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics