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Grade: 12th pass
        
A rigid and insulated tank of 3 m^3 volume is divided into two compartments. One compartment of volume 2 m^3 contain an ideal gas at 0.8314 Mpa and 400 K and while the second comprtmnt of volume 1 m^3 contains the same gas at 8.314 Mpa  and 500 K.If partition between the two compartments is ruptured what is the temperature of the gas. 
The correct option is 480 K... 
Kindly share the method to solve it
4 months ago

Answers : (2)

Vikas TU
7091 Points
							
No. of moles (n1) = P1V1/RT1 = (2*0.8314*10^6)/(0.8314*400*10^4) => ½
No. of moles (n2) = P2V2/RT2 = (1*0.8314*10^6)/(0.8314*10^4*500) = 2
 
Total moles = 2 + ½ = 5/2 = 2.5 moles. 
 
Since total moles are equal before and after reaction:
Peq. = (P1V1 + P2V2)/(P1 + P2) => ((0.831)*2 + (8.314)*1))*10^6/3 => 3.32 x 10^6 Pa.
 
Peq.*Vtotal  = ntotal*R*T
T = (3.32)*3*10^6/(2.5 * 8.314 * 10^6) => 480 K
4 months ago
Raj Shakya
16 Points
							
no. of moles  n1 = P1V1/RT1 = (2*0.8314)/(R*400)  ------i
no. of moles n2 = P2V2/RT2 = (1*8.314)/(R*500)  ----- ii
 
as container is insulated heat realsed from 1 compartment will be equal to heat absorbed by another compartment to attain a common temperature.
suppose C as molar heat capacity since C = dq/(dT*n)
so n1*C*(T – 400) =  – n2 *C*(T – 500)
so putting n1 and n2
2*0.8314*(T-400)/(R*400) = 1*8.314*(500-T)/(R*500)
by solving 5T = 2400
T =480
4 months ago
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