Question

Hi...

Two systems have the following equations of state and are contained in a closed cylinder, separated by a fixed, adiabatic and impermeable piston. N_1N1​ = 2 and N_2N2​ = 1.5 moles. The initial temperatures are T_1T1​ = 175 K and T_2T2​ = 400 K. The total volume is 0.025 m^33. The piston is allowed to move and heat transfer is allowed across the piston. Determine the final temperature of the system (in Kelvin).

1T1=32RN1U1 ,P1T1=RN1V1

and

1T2=52RN2U2 , P2T2=RN2V2

can you help me with solving this problem?

Answer 1

the problem is based on the concepts of thermodynamics.

we have to find the final temperature if the system.

when the piston will be allowed to moved, it will react at the equilibrium when , pressure on both side will be same, and the temperature on both side will be same because heat transfer is also happening.

so, the data given to us is,

1. moles = n1 = 2mol

2. moles = n2 = 1.5 mol

3. temperature = T1 = 175 K

4. temperature = T2 = 400 K

now the thermal equilibrium is archived when the sum of initial internal energy will be same as the sum of final internal energy of both system.

we are given that,

Cv of system 1 = 3/2R

and Cv of system 2 = 5/2R

so,

we know that internal energy of the systems is given as,

here

n = moles

Cv =specific heat at constant volume

T= temperature in Kelvin

so,

let,

U = initial internal energy of system 1

U2 = initial internal energy of system 2

U1' = final internal energy of system 1

U2' = final internal energy of system 2

so,the condition of thermal equilibrium will be,

U1 +U2 = U1' + U2'

so,

n1*3/2R*T1 + n2*5/2R*T2 = n1*3/2R*T' + n2*5/2R*T'

let T' be the final temperature of the system

so,,we get,

T' = (n1*3/2*T1 + n2*5/2*T2)/(n1*3/2 + n2*5/2)

so we substitute the values we get,

$T' = \frac{(2mol*3/2*175K) + (1.5mol*5/2*400K)}{(2mol*3/2)+(1.5mol*5/2)}$

we get

.........................ans.

hence ,

the final temperature of the system will be 300K.