Give step-by-step solution with explanation and final answer:
{give me the answer without using bullet-points}321, The state of an ideal gas with C= (S/R is changed from P = 1 bur and V} = 12 m*
10m

Question

Give step-by-step solution with explanation and final answer:
{give me the answer without using bullet-points}321, The state of an ideal gas with C= (S/R is changed from P = 1 bur and V} = 12 m*
10m 12 bar dV = | m by the following mechamscally reversible processes:
(@) Bother compression.
(6) Adibatic compression followed by cooing of constant presare.
(©) Adibatic compression followed by cooing af constant lume.
ud) Heating st constant von followed by onlin 5 comntan preva.
(e) Cooling at constant pressor followed by esting a constant vole.
Caleulae 0. W. AU and AH for each of these processes, ad sketch the paths of al
processes on a single PV diagram.

VIPSolutions · Accepted Answer

Let's break down the problem step by step.  
Given:  
- The gas is ideal, with $ C_P = \frac{5}{2} R $.
- Initial state: $ P_1 = 1 $ bar, $ V_1 = 12 $ m³.
- Final state: $ P_2 = 12 $ bar, $ V_2 = 1 $ m³.
- We are to consider four processes and calculate $ Q $ (heat), $ W $ (work), $ \Delta U $ (change in internal energy), and $ \Delta F $ (change in Helmholtz free energy).

Let's denote $ n $ as the number of moles and $ R $ as the gas constant.  
First, calculate the number of moles:

From the ideal gas law:
$ P_1 V_1 = nRT_1 $  
$ P_2 V_2 = nRT_2 $

First, find $ n $ in terms of $ T_1 $ or $ T_2 $:
$ n = \frac{P_1 V_1}{RT_1} = \frac{P_2 V_2}{RT_2} $

### 1. Isothermal Compression

For isothermal compression:
- $ T_1 = T_2 = T $
- $ \Delta U =  $ (since internal energy for an ideal gas depends only on temperature)
- Work done on the gas:  
  $ W = nRT \ln \frac{V_2}{V_1} $  
  Since $ V_2 < V_1 $, $ W $ will be negative (work done **on** the gas).
- The heat exchanged:  
  $ Q = W $ (since $ \Delta U =  $)
- Helmholtz Free Energy:  
  $ \Delta F = -nRT \ln \frac{V_2}{V_1} $

Now, $ nRT = P_1V_1 $ or $ P_2V_2 $ at the same temperature, so:
$ nRT = (1\,\text{bar})(12\,\text{m}^3) = (12\,\text{bar})(1\,\text{m}^3) $
Convert bar to SI units (1 bar = 100,000 Pa), but since ratios are used, units will cancel in logarithms.

$ W = nRT \ln \frac{1}{12} = nRT \ln(.0833) = -2.4849\, nRT $
Let $ nRT = P_1V_1 = 12 $ bar·m³

So, $ W = 12 \ln \frac{1}{12} = 12 \times (-2.4849) = -29.819 $ bar·m³

Convert to Joules: $ 1\,\text{bar}\cdot \text{m}^3 = 10^5\,\text{J} $
So, $ W = -2,981,900\,\text{J} $

$ Q = W $
$ \Delta U =  $
$ \Delta F = -W $ (since isothermal)

### 2. Adiabatic Compression, Then Cooling at Constant Pressure

**Step 1:** Adiabatic Compression:  
- No heat exchange: $ Q_1 =  $
- Use adiabatic relation: $ P_1 V_1^{\gamma} = P_a V_a^{\gamma} $, where $ \gamma = C_P/C_V = (5/2)R/[(5/2)R - R] = 5/3 $
- Find intermediate state ($ V_a $, $ P_a $) such that final volume $ V_2 = 1\,\text{m}^3 $, final pressure $ P_2 = 12\,\text{bar} $.

But since the process ends at $ P_2, V_2 $, the adiabatic ends at the volume where further cooling at constant pressure brings us to the final state.

For adiabatic:
$ P_1 V_1^{\gamma} = P_x V_x^{\gamma} $,  
Let $ V_x $ be the volume after adiabatic compression, where pressure is $ P_2 $.
$ P_1 V_1^{\gamma} = P_2 V_x^{\gamma} $
$ V_x = V_1 \left(\frac{P_1}{P_2}\right)^{1/\gamma} = 12 \left(\frac{1}{12}\right)^{3/5} $

$ \frac{1}{12}^{.6} \approx .164 $
$ V_x = 12 \times .164 = 1.97\,\text{m}^3 $

So, after adiabatic compression:  
$ P = 12\,\text{bar}, V = 1.97\,\text{m}^3 $

**Step 2:** Cooling at constant pressure to $ V_2 = 1\,\text{m}^3 $  
At constant pressure, cooling from $ V_x $ to $ V_2 $.

Calculate $ \Delta U $:  
Since $ \Delta U = nC_V \Delta T $, and $ T = \frac{PV}{nR} $,  
$ \Delta U = nC_V (T_2 - T_1) $, but need to find $ T_1 $ and $ T_2 $:

- $ T_1 = \frac{P_1 V_1}{nR} $
- $ T_2 = \frac{P_2 V_2}{nR} $

Let $ nR = P_1V_1/T_1 = 1 \times 12 / T_1 $
But for ratios, just use the proportionality:

$ \frac{T_2}{T_1} = \frac{P_2 V_2}{P_1 V_1} = \frac{12 \times 1}{1 \times 12} = 1 $

So $ T_2 = T_1 $, but that's not possible, let's check at each step:

After adiabatic, $ T_x = \frac{P_2 V_x}{nR} $
Initial, $ T_1 = \frac{P_1 V_1}{nR} $

$ \Delta U = nC_V (T_2 - T_1) $
$ \Delta U = nC_V \left(\frac{P_2 V_2 - P_1 V_1}{nR}\right) $
$ C_V = C_P - R = \frac{5}{2}R - R = \frac{3}{2}R $
$ \Delta U = \frac{3}{2}(P_2 V_2 - P_1 V_1) $

Plugging in:  
$ P_1 V_1 = 12 $, $ P_2 V_2 = 12 $, so $ \Delta U = \frac{3}{2}(12 - 12) =  $

**But this suggests $ \Delta U =  $ for all paths, which is only true for isothermal.**  
Let me clarify for the adiabatic step.

For adiabatic step:  
$ T_x = T_1 \left(\frac{V_1}{V_x}\right)^{\gamma-1} $
$ T_1 = \frac{P_1 V_1}{nR} $
$ T_x = T_1 \left(\frac{12}{1.97}\right)^{2/3} $

Calculate $ \frac{12}{1.97} \approx 6.09 $, $ 6.09^{2/3} \approx 3.3 $
So $ T_x = 3.3 T_1 $

Now, at constant pressure, volume decreases from $ V_x = 1.97 $ to $ V_2 = 1 $,  
So $ T_2 = T_x \frac{V_2}{V_x} = 3.3 T_1 \times \frac{1}{1.97} = 1.675 T_1 $

So overall,  
$ \Delta U = nC_V (T_2 - T_1) = nC_V (1.675 T_1 - T_1) = nC_V (.675 T_1) $
Recall $ nRT_1 = P_1 V_1 = 12 $; $ C_V = 3/2 R $

$ n = \frac{P_1 V_1}{RT_1} $
Thus,
$ \Delta U = \frac{P_1 V_1}{RT_1} \cdot \frac{3}{2} R \cdot .675 T_1 = 12 \cdot \frac{3}{2} \cdot .675 = 12 \cdot 1.0125 = 12.15 $ bar·m³

Convert to Joules: $ 12.15 \times 10^5 = 1,215,000 $ J

Work done:

- **Adiabatic step:**  
$ W_1 = \frac{P_2 V_x - P_1 V_1}{1 - \gamma} $
$ \gamma = 5/3 $, $ 1 - \gamma = -2/3 $
$ W_1 = \frac{12 \times 1.97 - 1 \times 12}{-2/3} = \frac{23.64 - 12}{-2/3} = \frac{11.64}{-2/3} = -17.46 $ bar·m³

- **Isobaric cooling:**  
\( W_2 = P_2 (V

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