Question

Two systems have the following equations of state and are contained in a closed cylinder, separated by a fixed, adiabatic and impermeable piston. N_1N 1
​ = 2 and N_2N 2
​ = 1.5 moles. The initial temperatures are T_1T 1
​ = 175 K and T_2T 2
​ = 400 K. The total volume is 0.025 m^3 3
. The piston is allowed to move and heat transfer is allowed across the piston. Determine the final temperature of the system (in Kelvin).

Answer 1

Answer:

Given:

N1= 2 and N2 = 1.5 moles.

The initial temperatures are T1 = 175 K and T2 = 400 K.

The total volume is 0.025 m^3.

1/T1=(3/2)R*N1/U1 ,P1/T1=R*N1/V1

1/T2=(5/2)R*N2/U2 , P2/T2=R*N2/V2

To find:

Determine the final temperature of the system (in Kelvin).

Solution:

Let the volumes of two systems be, V1 and V2

From given, we have,

V1 + V2 = V = 0.025 m³

As given, that both systems are in a closed container in adiabatic condition, we have,

T1 = 175 K

T2 = 400 K

w.k.t PV = nRT

nRT1/P1 + nRT2/P2 = nRT/P

2R×175/P + 1.5R×400/P = 3.5RT/P

2 × 175 + 1.5 × 400 = 3.5T

350 + 600 = 3.5T

950 = 3.5T

T = 950/3.5

∴T = 271.42 K

Hence the final temeperature.

Answer 2

Answer:

271.42 K

Explanation:

Temparature

271 K