Question

11. A closed, diatomic gas system, m=0.025kg, has a pressure, P=1.35x105Pa and temperature, T=390.OK. The average mass per gas particle is 6.63x10-26kg. At constant volume, the temperature of the gas is then decreased to 275.OK. a. Calculate Vems for the gas at the end of the cooling process. b. Determine Q for this process. c. Determine the specific heat capacity of the gas under these conditions.

Answer 1

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T- w N 25 6.63 x 1023 Here, m = 0.025 Kg P = 1,35 * 105 Pa 390 K Average mass of gassparticle, mape m = 6.63 x 1826 kg Numer of molecule/ gas particle in the system 0.025 mi 6.63 x 10 •26 Number of moles of gas in the syetem mole NA 6.63 6.022 1023) = 0.626 moles Final temperature of gas, T = 275.0k 3KT' Vү ум е 3x 1.38 x H 10²3 x 275 mi 6.63 x 10-26 X1022 x n Vrms 414.4 mis (b) x -x8.31 Gas releases the heat at constant volume. Q change in internal energy = ng AT (ER) (T-T for diatomic gas, Ev - 5 R) 10.626) -31) 275 - 390) Q = - 1495.6 J Negative sign shows that heat in released specific hest capacity, s = ? Q = ma mSAT ancy AT nly (0.626) (C) Here X V 0.025 s = 520.21 I kgl kl

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