(A) A cylinder with a movable piston contains gas at a temperature of 28.0°C, a

Question

(A) A cylinder with a movable piston contains gas at a temperature of 28.0°C, a volume of 1.90 m3, and an absolute pressure of 0.200 multiplied by 105 Pa. What will be its final temperature if the gas is compressed to 0.900 m3 and the absolute pressure increases to 0.800 multiplied by 105 Pa?

(B)Lead pellets, each of mass 1.50 g, are heated to 200°C. How many pellets must be added to 450 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C? Neglect any energy transfer to or from the container.
pellets

Explanation / Answer

(A) Given Data:        Initial temperature, T1 = 280 C                                             = 301 K        Initial Volume , V1 = 1.90 m3        Initial Pressure, P1 = 0.2 *105 Pa        Final pressure, P2 = 0.8*105 Pa        Final Volume, V2 = 0.900 m3 From ideal gas law,                                                PV = n RT                                       P1 V1 / T1 = P2 V2 / T2 ( 0.2 *105 Pa) (1.90 m3) / (301 K) = (0.8*105 Pa) (0.900 m3) / T2                                                 T2    = 570.5 K or 297.50 C ------------------------------------------------------------------------------------- (B) Let number of lead pellets be n        Mass of each lead pellet is, m = 1.50 g                                                        = (1.50 g ) (10-3 kg /1 g)                                                        = 0.0015 kg       Mass of water , M = 450 g                                     = (450 g ) (10-3 kg /1 g)                                     = 0.45 kg       Initial temperature of lead, Til = 2000 C       Initial temperature of water , Tiw = 200 C      Equilibrium temperature, T = 250 C      Specific heat of lead , CL = 130 J / kg0 C      Specific heat of water , Cw = 4186 J/ kg0 C Solution:     Heat lost by lead = Heat gained by water     n (m) CL ( Til- T) = M Cw (T - Tiw)    n (0.0015 kg) (130 J / kg0 C) ( 2000 C- 250 C) = (0.45 kg) (4186 J/ kg0 C) ( 250 C - 200 C)                                      n = 276 ------------------------------------------------------------------------------------- (B) Let number of lead pellets be n        Mass of each lead pellet is, m = 1.50 g                                                        = (1.50 g ) (10-3 kg /1 g)                                                        = 0.0015 kg       Mass of water , M = 450 g                                     = (450 g ) (10-3 kg /1 g)                                     = 0.45 kg       Initial temperature of lead, Til = 2000 C       Initial temperature of water , Tiw = 200 C      Equilibrium temperature, T = 250 C      Specific heat of lead , CL = 130 J / kg0 C      Specific heat of water , Cw = 4186 J/ kg0 C Solution:     Heat lost by lead = Heat gained by water     n (m) CL ( Til- T) = M Cw (T - Tiw)    n (0.0015 kg) (130 J / kg0 C) ( 2000 C- 250 C) = (0.45 kg) (4186 J/ kg0 C) ( 250 C - 200 C)                                      n = 276                                     = 0.45 kg       Initial temperature of lead, Til = 2000 C       Initial temperature of water , Tiw = 200 C      Equilibrium temperature, T = 250 C      Specific heat of lead , CL = 130 J / kg0 C      Specific heat of water , Cw = 4186 J/ kg0 C Solution:     Heat lost by lead = Heat gained by water     n (m) CL ( Til- T) = M Cw (T - Tiw)    n (0.0015 kg) (130 J / kg0 C) ( 2000 C- 250 C) = (0.45 kg) (4186 J/ kg0 C) ( 250 C - 200 C)                                      n = 276

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