Three moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 430 to 160 K.

(a) What is the work done by (or done to) the gas? Include the algebraic sign.
J

(b) What is the change in the internal energy of the gas? Include the algebraic sign.
J

To calculate the work done by (or done to) the gas during an adiabatic process, we can use the formula:

W = (Cv - Cv') x ΔT

Where:
W is the work done by (or done to) the gas,
Cv is the molar heat capacity at constant volume,
Cv' is the molar heat capacity at constant pressure,
and ΔT is the change in temperature.

For a monatomic ideal gas, the molar heat capacity at constant volume (Cv) is given by:

Cv = (3/2)R

Where R is the ideal gas constant.

(a) To find the work done by (or done to) the gas, we need to calculate the molar heat capacity at constant pressure (Cp). Since the gas is adiabatically expanding, there is no heat exchange with the surroundings, so ΔQ is zero. Therefore, Cv = Cp.

Cv' = Cp = (5/2)R

Using the given information, we can calculate the work done by (or done to) the gas:

ΔT = 160 K - 430 K = -270 K

W = (Cv - Cv') x ΔT
W = [(3/2)R - (5/2)R] x (-270)
W = -2R x (-270)
W = 540R

The work done by (or done to) the gas is 540R J.

(b) The change in internal energy (ΔU) during an adiabatic process can be calculated using the formula:

ΔU = (Cv - Cv') x ΔT

Using the given information, we can calculate the change in internal energy:

ΔU = (Cv - Cv') x ΔT
ΔU = [(3/2)R - (5/2)R] x (-270)
ΔU = -2R x (-270)
ΔU = 540R

The change in internal energy of the gas is 540R J.

To calculate the work done by (or done to) the gas in an adiabatic expansion, we can use the formula:

W = -(γ / (γ - 1)) * P₁V₁ * (V₂/V₁)^γ - P₂V₂

where W is the work done, γ is the heat capacity ratio (which is 5/3 for a monatomic ideal gas), P₁ and P₂ are the initial and final pressures, V₁ and V₂ are the initial and final volumes.

To calculate the change in the internal energy of the gas, we can use the formula:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat transferred, and W is the work done.

However, in this case, we are given that the process is adiabatic, meaning no heat is transferred (Q = 0). Therefore, the change in internal energy simplifies to just the work done:

ΔU = W

Now let's calculate the value for the work done and the change in internal energy.

(a) Work done:

First, we need to convert the temperatures from Celsius to Kelvin:

T₁ = 430 K
T₂ = 160 K

Next, we can use the ideal gas law to relate the initial and final pressures and volumes:

P₁V₁ / T₁ = P₂V₂ / T₂

Since the number of moles (n) and gas constant (R) are constant, we can ignore them:

P₁V₁ = P₂V₂

Therefore,

P₂ = P₁ * (V₁ / V₂)

Now, let's substitute the values into the work formula:

W = -(γ / (γ - 1)) * P₁V₁ * (V₂/V₁)^γ - P₂V₂

W = -(5/3) / (5/3 - 1) * P₁V₁ * (V₂/V₁)^(5/3) - P₁ * (V₁ / V₂) * V₂

W = -(5/3) / (2/3) * P₁V₁ * (V₂/V₁)^(5/3) - P₁V₁

(b) Change in internal energy:

Since the process is adiabatic (Q = 0), the change in internal energy is equal to the work done:

ΔU = W

Substituting the value of work done from above:

ΔU = -(5/3) / (2/3) * P₁V₁ * (V₂/V₁)^(5/3) - P₁V₁

Now, we need to evaluate these expressions by plugging in the given values of the initial and final temperatures, and any other known variables from the problem.

Do you have the values of the initial and final volumes or any other information that could help solve this problem?

dE = T dS - PdV

Adiabatic change: Heat absorbed T dS = 0, the change in internal energy is due to work
P dV performed by the gas.

Monoatomic gas:

E = 3/2 N k T

1 mole = 6.022*10^23

k = 1.38*10^(-23)J/K

So, the temperature change gives you know Delta E (which is clearly negative) and that equals minus the work done by the gas (this is positive as work done by the gas comes at the expense of the gas)