Given:
Pressure (P1) = 4 atm
Pressure (P2) = 2 atm
Temperature (T) = 300 K

Calculating q:
Since the process is reversible and isothermal, the temperature remains constant throughout the process. Therefore, the change in internal energy (∆E) is zero, and q = ∆E + P∆V. As ∆E is zero, q = P∆V.

To calculate q, we need to determine the change in volume (∆V). According to Boyle's Law, P1V1 = P2V2. Rearranging the equation, we get ∆V = V2 - V1 = V1(P2/P1 - 1).

Substituting the given values, ∆V = V1(2/4 - 1) = -V1/2.

Since the gas is expanding, the change in volume is positive. Therefore, ∆V = V1/2.

Now, we can calculate q using q = P∆V. Substituting the values, q = (4 atm)(V1/2) = 2V1 atm.

Therefore, q = 2V1 atm.

Calculating w:
In a reversible isothermal process, the work done (w) by the gas is given by the equation w = -P∆V.

Substituting the values, w = -(4 atm)(V1/2) = -2V1 atm.

Therefore, w = -2V1 atm.

Calculating ∆E:
Since the temperature remains constant during the process, ∆E = 0.

Therefore, ∆E = 0.

Calculating ∆H:
Since the process is reversible and isothermal, ∆H = q.

Substituting the value of q, ∆H = 2V1 atm.

Therefore, ∆H = 2V1 atm.

Summary:
- q = 2V1 atm
- w = -2V1 atm
- ∆E = 0
- ∆H = 2V1 atm