1.We have a container of a hot ideal monatomic gas. The volume of the container

Question

1.We have a container of a hot ideal monatomic gas. The volume of the container is 25 liters. The temperature of the gas is 455°C, and its pressure is 0.858 × 105 Pa. We allow the gas to cool down to room temperature, which at the time is 21°C. We do not allow the volume of the gas to change. (a) Find the final pressure (Pa) of the gas. (b) Find the amount of heat (J) that passed from the gas to its surroundings as it cooled. (Give a positive value.) (c) Find the change of internal energy (J) of the gas.

Explanation / Answer

a) As volume and no of moles remains constant of our gas P/T = CONSTANT (P1/T1)=(P2/T2) => (21+273)*(0.858*(10^5)/(455+273)) = P2 P2=FINAL PRESSURE OF THE GAS= 0.3465x10^5 Pa B) From first law of thermodynamics delta U = delta Q -delta W Now as volume doesnt changes thus there is no work by system on surrounding. hence, delta W= 0 b)Thus deltaQ = delta U = (F/2)nR*(deltaT) In case of monoatomic gas, F=3 [DEGREE OF FREEDOM] Thus deltaQ = delta U = (1.5)*25*(10^(-3))*(0.858*(10^5))*(21-455)/(455+273) =-1 918.125 [Thus this is the energy gained by system] Thus energy gained by surrounding= -deltaQ = 1918.125 J = 1.918 kJ c) And change in internal energy = deltaU = -1.918 kJ

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