Question

One mole of ideal gas goes through the following transformations:

Step I: Isochoric cooling to 1/3 of the initial temperature.

Step II: Adiabatic compression to its initial pressure.

Step III: Isobaric expansion back to the initial state.

If the initial temperature and pressure are 600 K and 3.00 atm, and if the ideal gas is monoatomic ($C_v$ = 3/2R), what is the work for the third step (going from state 3 to state 1)?

Answer 1

2408.06 J

Answer 2

The problem describes a thermodynamic cycle for one mole of an ideal monoatomic gas. We need to calculate the work done during the third step, which is an isobaric expansion from state 3 to state 1.

Given:

• Number of moles, $n = 1$ mol
• Initial temperature, $T_1 = 600$ K
• Initial pressure, $P_1 = 3.00$ atm
• For a monoatomic gas, $C_v = \frac{3}{2}R$.
• Also, $C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R$.
• The adiabatic index, $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.

Let's analyze the states and transformations:

State 1 (Initial State): $P_1 = 3.00$ atm $T_1 = 600$ K Using the ideal gas law, $P_1V_1 = nRT_1$: $V_1 = \frac{nRT_1}{P_1} = \frac{1 \times R \times 600}{3.00} = 200R$

Step I: Isochoric cooling from State 1 to State 2 This means the volume remains constant: $V_2 = V_1 = 200R$. The temperature cools to 1/3 of the initial temperature: $T_2 = \frac{1}{3}T_1 = \frac{1}{3} \times 600 \text{ K} = 200 \text{ K}$. For an isochoric process, $\frac{P_1}{T_1} = \frac{P_2}{T_2}$: $P_2 = P_1 \frac{T_2}{T_1} = 3.00 \text{ atm} \times \frac{200 \text{ K}}{600 \text{ K}} = 3.00 \text{ atm} \times \frac{1}{3} = 1.00 \text{ atm}$.

State 2: $P_2 = 1.00$ atm $T_2 = 200$ K $V_2 = 200R$

Step II: Adiabatic compression from State 2 to State 3 The gas is compressed to its initial pressure: $P_3 = P_1 = 3.00$ atm. For an adiabatic process, the relation between temperature and pressure is $T P^{(1-\gamma)/\gamma} = \text{constant}$, or $T^\gamma P^{1-\gamma} = \text{constant}$. So, $T_2^\gamma P_2^{1-\gamma} = T_3^\gamma P_3^{1-\gamma}$. Rearranging for $T_3$: $T_3 = T_2 \left(\frac{P_3}{P_2}\right)^{(\gamma-1)/\gamma}$ Substitute the values: $\gamma = 5/3$, $\gamma-1 = 2/3$, so $\frac{\gamma-1}{\gamma} = \frac{2/3}{5/3} = \frac{2}{5}$. $T_3 = 200 \text{ K} \times \left(\frac{3.00 \text{ atm}}{1.00 \text{ atm}}\right)^{2/5}$ $T_3 = 200 \times (3)^{2/5}$ K

State 3: $P_3 = 3.00$ atm $T_3 = 200 \times (3)^{2/5}$ K Using the ideal gas law, $P_3V_3 = nRT_3$: $V_3 = \frac{nRT_3}{P_3} = \frac{1 \times R \times (200 \times 3^{2/5})}{3.00} = \frac{200}{3} R \times 3^{2/5}$

Step III: Isobaric expansion from State 3 to State 1 This step returns the gas to its initial state (State 1). Since $P_3 = P_1 = 3.00$ atm, this is an isobaric process. The work done by the gas in an isobaric process is given by $W = P \Delta V$. $W_{31} = P_1 (V_1 - V_3)$ Substitute $V = \frac{nRT}{P}$: $W_{31} = P_1 \left(\frac{nRT_1}{P_1} - \frac{nRT_3}{P_1}\right)$ Since $P_1$ is constant during this step, it simplifies to: $W_{31} = nR(T_1 - T_3)$ Substitute the known values for $n$, $T_1$, and $T_3$: $W_{31} = 1 \times R \times (600 - 200 \times 3^{2/5})$ $W_{31} = R (600 - 200 \times 3^{2/5})$

To get a numerical value, we use $R = 8.314 \text{ J/mol K}$. First, calculate $3^{2/5}$: $3^{2/5} = (3^2)^{1/5} = 9^{1/5} \approx 1.55184557$ Now, substitute this value into the expression for $W_{31}$: $W_{31} = 8.314 \times (600 - 200 \times 1.55184557)$ $W_{31} = 8.314 \times (600 - 310.369114)$ $W_{31} = 8.314 \times 289.630886$ $W_{31} \approx 2408.06$ J

The work done for the third step (isobaric expansion) is positive, which is consistent with work done by the gas during expansion.