Two thermally insulated vessels are connected by a narrow tube fitted with a val
ID: 2019823 • Letter: T
Question
Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure below. One vessel of volume V1 = 17.0 L, contains oxygen at a temperature of T1 = 295 K and a pressure of P1 = 1.76 atm. The other vessel of volume V2 = 22.5 L contains oxygen at a temperature of T2 = 450 K and a pressure of P2 = 2.65 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure?Explanation / Answer
Moles of the oxygen gas in Vessel1=P1V1/RT1
=(1.76*1.013*10^5 Pa)*(17*10^-3 m^3)/(8.314*295)
n1 =1.235 moles
Similarly moles of the oxygen gas in Vessel2=P2V2/RT2
=(22.5*10^-3 m^3)*(2.65*1.013*10^5 Pa)/(8.314*450)
n2 =1.614 moles
A) so the equilibrium temperature of the gas
Heat lost=heat gained
n1M1*c1(Tf-295)= n2M2*c2(450-Tf) [Because change in heat is mcT]
As M1=M2 [Because both the gases are Oxygen]
Therefore 1.235*(Tf-295)+1.614(Tf-450) = 0
2.849 Tf =1090.625
Thats gives Tf=382.80 K
B)Pressure of the final mixture =nRT/V=(2.849*8.314*382.80)/(17+22.5)*10^-3 m^3
=229550.00=2.29 atm
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