Ch01 Thermodynamics

PHYSICS 2ND PAPER CHAPTER 1 Thermodynamics Question 1 [R. B. 2017] Some gas is confined in a cylinder fitted with a piston. When 600 J of heat is slowly supplied to the gas at a constant pressure of 300 Pa, 900 J of work is done by the system. (a) Find the change in volume of the gas. (b) “According to the stem the principle of conservation of energy is not violated”. Justify its correctness with mathematical analysis. Solution (a) Here, Constant pressure 𝑃= 300 Pa = 300 N m−2 Work done by the gas 𝑊= 900 J At constant pressure, the work done by a gas is 𝑊= 𝑃Δ𝑉. So Δ𝑉= 𝑊 𝑃 = 900 J 300 N m−2 = 3 m3 Δ𝑉= 3 m3 (b) Here, Heat supplied to the gas Δ𝑄= 600 J Work done by the gas 𝑊= 900 J The first law of thermodynamics is the principle of conservation of energy for heat and work: Δ𝑄= Δ𝑈+ 𝑊 So the change in internal energy of the gas is Δ𝑈= Δ𝑄−𝑊 = 600 J −900 J = −300 J The negative sign shows that the internal energy of the gas decreases by 300 J. ahbab.dev 1

Energy balance: Energy taken in = heat supplied + decrease in internal energy = 600 J + 300 J = 900 J Energy given out = work done by the gas = 900 J The gas does 900 J of work. It gets 600 J from the supplied heat and the remaining 300 J from its own internal energy. No energy is created or destroyed. Conclusion: the statement is correct, because the principle of conservation of energy is not violated. Δ𝑈= −300 J, energy taken in = energy given out = 900 J ahbab.dev 2

Question 2 [R. B. 2019] Sujan and Shaily, two students of class twelve, halved the volume of 1 mol of a gas at 27 ◦C temperature and 300 cm of mercury pressure by isothermal and adiabatic processes respectively. The gas is diatomic (𝛾= 1.4, 𝑅= 8.314 J mol−1 K−1). (a) What will be the temperature of the gas after the thermodynamic change done by Shaily? (b) In the light of the stem, between Sujan and Shaily, who will perform more work? Explain with mathematical analysis. Solution (a) Here, for Shaily’s adiabatic compression: Initial temperature 𝑇1 = (27 + 273) K = 300 K Volume ratio 𝑉1/𝑉2 = 2 Ratio of specific heats 𝛾= 1.4 For an adiabatic process, 𝑇𝑉𝛾−1 remains constant. So 𝑇1𝑉𝛾−1 1 = 𝑇2𝑉𝛾−1 2 𝑇2 = 𝑇1 𝑉1 𝑉2 𝛾−1 = 300 × 21.4−1 = 300 × 20.4 = 300 × 1.3195 = 395.85 K In degree Celsius, 𝑡2 = 395.85 −273 = 122.85 ◦C. 𝑇2 = 395.85 K = 122.85 ◦C (b) Here, 𝑛= 1 mol, 𝑇1 = 300 K, 𝑅= 8.314 J mol−1 K−1, 𝛾= 1.4 and 𝑉2 = 𝑉1/2. 1. Sujan (isothermal compression): 𝑊Sujan = 𝑛𝑅𝑇1 ln 𝑉2 𝑉1 = 1 × 8.314 × 300 × ln(0.5) = 2494.2 × (−0.6931) = −1728.8 J 2. Shaily (adiabatic compression): From (a), 𝑇2 = 395.85 K. 𝑊Shaily = 𝑛𝑅(𝑇1 −𝑇2) 𝛾−1 = 1 × 8.314 × (300 −395.85) 1.4 −1 = 8.314 × (−95.85) 0.4 = −1992.3