The temperature of 4.60 mol of an ideal monatomic gas is raised 15.0 K in an adiabatic process. What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change ΔEint in internal energy of the gas, and (d) the change ΔK in the average kinetic energy per atom?

Since no heat is added, you must be increasing the temperature by isentropic compression. Negative work is done BY the gas.

For (a) Calculate the P dV integral using the fact that
Po*Vo^(5/3) = P*V^(5/3)

Work = integral P dV =
Po*Vo^(5/3)* integral of dV/V^(5/3)

More information is required about the initial conditions of the gas. Does it start out at S.T.P. ?

(b) Since this is an adiabatic process, the heat transferred in or out is zero. That is the definition of adiabatic.

To solve this problem, we need to use the following equations:

(a) The work done by the gas in an adiabatic process is given by:
W = -∆Eint

(b) The energy transferred as heat in an adiabatic process is zero:
Q = 0

(c) The change in internal energy of the gas is given by:
∆Eint = nCv∆T

(d) The change in the average kinetic energy per atom is given by:
∆K = (3/2)nR∆T

Now let's calculate each value using the given information:

(a) The work done by the gas:
W = -∆Eint
From equation (c), ∆Eint = nCv∆T
Here, n = 4.60 mol (number of moles of gas)
Cv = 3/2 R (molar specific heat capacity at constant volume for a monatomic gas)
∆T = 15.0 K (change in temperature)

Therefore,
W = -nCv∆T
= -(4.60 mol) * (3/2 R) * (15.0 K)
= -69.0 R K

(b) The energy transferred as heat:
Q = 0 (because the process is adiabatic)

(c) The change in internal energy of the gas:
∆Eint = nCv∆T
From equation (c), ∆Eint = nCv∆T
Here, n = 4.60 mol (number of moles of gas)
Cv = 3/2 R (molar specific heat capacity at constant volume for a monatomic gas)
∆T = 15.0 K (change in temperature)

Therefore,
∆Eint = (4.60 mol) * (3/2 R) * (15.0 K)
= 103.5 R K

(d) The change in the average kinetic energy per atom:
∆K = (3/2)nR∆T
Here, n = 4.60 mol (number of moles of gas)
R (universal gas constant) = 8.314 J/(mol K)
∆T = 15.0 K (change in temperature)

Therefore,
∆K = (3/2)(4.60 mol)(8.314 J/(mol K))(15.0 K)
= 174.9 J

So, the answers are:
(a) The work done by the gas is -69.0 R K.
(b) The energy transferred as heat is 0.
(c) The change in internal energy of the gas is 103.5 R K.
(d) The change in the average kinetic energy per atom is 174.9 J.

To answer these questions, we need to make use of the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, it can be written as:

ΔEint = Q - W

Let's tackle each part of the question one by one:

(a) The work done by the gas (W):

In an adiabatic process, no heat is transferred between the system and its surroundings. Therefore, the work done is equal to the change in internal energy. So, W = ΔEint.

(b) The energy transferred as heat (Q):

In an adiabatic process, as mentioned earlier, there is no heat transfer. So, Q = 0.

(c) The change in internal energy (ΔEint):

The change in internal energy of an ideal gas can be calculated using the equation:

ΔEint = (3/2) × n × R × ΔT

Where:
- n is the number of moles of the gas (given as 4.60 mol)
- R is the molar gas constant (8.314 J/(mol·K))
- ΔT is the change in temperature (given as 15.0 K)

Using these values, we can calculate ΔEint.

ΔEint = (3/2) × 4.60 mol × 8.314 J/(mol·K) × 15.0 K

(d) The change in average kinetic energy per atom (ΔK):

The change in average kinetic energy per atom can be calculated using the equation:

ΔK = (3/2) × R × ΔT / N

Where:
- R is the gas constant as before
- ΔT is the change in temperature as before
- N is the number of atoms in the gas. In an ideal monatomic gas, there are Avogadro's number (6.022 × 10^23) of atoms per mole.

Using these values, we can calculate ΔK.

ΔK = (3/2) × 8.314 J/(mol·K) × 15.0 K / (6.022 × 10^23)

Now, you can substitute the given values into these equations and calculate the answers to parts (a), (c), and (d). Remember that part (b) is already given as Q = 0 in an adiabatic process.