The work done by one mole of a monatomic ideal gas (y=5/3) in expanding adiabatically is 825 J. The initial temperature and volume of the gas are 393 K and 0.100 m3. Obtain (a) the final temperature and (b) the final volume of the gas.

To solve this problem, we can use the following equations for adiabatic expansion of an ideal gas:

(a) The equation relating the initial and final temperatures (T₁ and T₂) and the gas constant (R):

T₁ / T₂ = (V₂ / V₁)^(y-1),

where y = 5/3 is the heat capacity ratio.

(b) The equation relating the initial and final volumes (V₁ and V₂) and the gas constant (R):

(V₂ / V₁) = (T₁ / T₂)^(1/(y-1)).

Now let's plug in the given values and solve the equations step-by-step:

Given data:
T₁ = 393 K
V₁ = 0.100 m³
y = 5/3
W = 825 J (work done by the gas)

(a) To find the final temperature (T₂):

We know that the work done by the gas (W) is given by: W = (y / (y-1)) * R * (T₁ - T₂),

We need to find R, the gas constant. Since we are given that the gas is monatomic, the gas constant (R) is given as R = (y / (y-1)) * n * Cv, where n is the number of moles and Cv is the molar heat capacity at constant volume.

For a monatomic ideal gas, Cv = (3 / 2) * R, so we can substitute this into the equation:

R = (y / (y-1)) * n * Cv = (5/3) / (5/3 - 1) * n * (3/2) * R = 3n * R,

where n = 1 mole, so R = (8/3) * R.

Substituting this value of R into the equation for work done, we get:

825 J = ((5/3) / ((5/3) - 1)) * (8/3) * R * (393 - T₂),

Now we can solve for T₂:

825 J = (5/3) * (8/3) * R * (393 - T₂),

825 J = (40/9) * (8/3) * R * (393 - T₂).

To find T₂, we can rearrange the equation:

825 J = (320/9) * R * (393 - T₂),

Dividing both sides of the equation by (320/9) * R:

825 J / ((320/9) * R) = 393 - T₂,

Substituting the value of R = (8/3) * R:

825 J / ((320/9) * (8/3) * R) = 393 - T₂,

825 J / (8533.33 R) = 393 - T₂,

825 / 8533.33 = 393 - T₂,

0.096646 = 393 - T₂,

T₂ = 393 - 0.096646,

T₂ ≈ 392.903 K.

Therefore, the final temperature (T₂) is approximately 392.903 K.

(b) To find the final volume (V₂):

Using the equation (V₂ / V₁) = (T₁ / T₂)^(1/(y-1)) and the values of T₁, T₂, V₁, and y:

(V₂ / 0.100) = (393 / 392.903)^(1/(5/3 - 1)),

(V₂ / 0.100) = (393 / 392.903)^(3/2),

(V₂ / 0.100) = (1.00022)^(3/2),

Taking both sides to the power of 2/3:

(V₂ / 0.100)^(2/3) = 1.00022,

V₂ / 0.100 = 1.00022^(3/2),

V₂ = (1.00022^(3/2)) * 0.100,

V₂ ≈ 0.100022 m³.

Therefore, the final volume (V₂) is approximately 0.100022 m³.

In summary:
(a) The final temperature (T₂) ≈ 392.903 K.
(b) The final volume (V₂) ≈ 0.100022 m³.