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_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​

Answer 1

Given that,

Total volume = 0.025 m³

Initial temperature = 175 K

Final temperature = 400 K

$N_{1}=2$

$N_{2}=1.5$

Let V₁ and V₂ are the volume of two systems.

We need to calculate the final temperature of the system

Using formula of volume

$V_{1}+V_{2}=V$

$\dfrac{nRT_{1}}{P_{1}}+\dfrac{nRT_{2}}{P_{2}}=\dfrac{nRT}{P}$

Put the value into the formula

$\dfrac{2R\times175}{P}+\dfrac{1.5R\times400}{P}=\dfrac{3.5RT}{P}$

$350+600=3.5T$

$T=\dfrac{350+600}{3.5}$

$T=271.4\ K$

Hence, The final temperature of the system is 271.4 K

Answer 2

Answer:

300

Explanation:

U1=1.5*N1*R*T1

U2=2.5*N2*R*T2

change in U1=change in U2

1.5*N1*R*(T-T1)=2.5*N2*R*(T2-T)

substitute values we get

T=4050/13.5=300K