Question

Consider the following volume−temperature (V−T) diagram for the expansion of 5 moles of an ideal monoatomic gas.

Considering only P-V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z is ______.
[Use the given data: Molar heat capacity of the gas for the given temperature range, C_V,_m = 12 J K⁻¹mol⁻¹ and gas constant, R=8.3 JK⁻¹ mol⁻¹]

Answer 1

Correct Answer: 8120

To solve the problem, we need to find the total change in enthalpy for the transformation of state in the sequence $X \to Y \to Z$.

1. Given Data:
We are given $n = 5$ moles of an ideal monoatomic gas.
Also, $C_{v,m} = 12 \, \text{J K}^{-1} \, \text{mol}^{-1}$ and $R = 8.3 \, \text{J K}^{-1} \, \text{mol}^{-1}$.

2. Relationship Between Enthalpy and Temperature:
For an ideal gas, enthalpy is a function of temperature only. The differential change in enthalpy is given by:

$ dH = n C_p dT$
where $C_p$ is the molar heat capacity at constant pressure.

3. Heat Capacities for a Monoatomic Gas:
For a monoatomic gas, we have:

$ C_v = \frac{3}{2} R$ and $C_p = \frac{5}{2} R$.

4. Calculating Molar Heat Capacity at Constant Pressure:
The molar heat capacity at constant pressure is:

$ C_{p,m} = C_{v,m} + R = 12 + 8.3 = 20.3 \, \text{J K}^{-1} \, \text{mol}^{-1}$.

5. Change in Enthalpy Formula:
The change in enthalpy is given by:

$ \Delta H = n C_p \Delta T$
or

$ \Delta H = n C_{p,m} \Delta T$.

6. Temperatures at States:
The temperature at state $X$ is $T_X = 335 \, \text{K}$, at state $Y$ is $T_Y = 335 \, \text{K}$, and at state $Z$ is $T_Z = 415 \, \text{K}$.

7. Change in Enthalpy for Process $X \to Y$:
Since the temperature is constant between states $X$ and $Y$, the change in enthalpy is:

$ \Delta H_{XY} = 0$.

8. Change in Enthalpy for Process $Y \to Z$:
The change in temperature for the process $Y \to Z$ is:

$ \Delta T = T_Z - T_Y = 415 - 335 = 80 \, \text{K}$.

9. Calculating the Change in Enthalpy for $Y \to Z$:
We can now calculate the change in enthalpy for the process $Y \to Z$ as:

$ \Delta H_{YZ} = n C_{p,m} \Delta T = 5 \times 20.3 \times 80 = 5 \times 1624 = 8120 \, \text{J}$.

10. Total Change in Enthalpy:
The total change in enthalpy is the sum of the changes for each process:

$ \Delta H = \Delta H_{XY} + \Delta H_{YZ} = 0 + 8120 = 8120 \, \text{J}$.

Final Answer:
The final answer is $\boxed{8120}$.

Answer 2

To solve the problem, we calculate the total change in enthalpy (\(\Delta H\)) for the process \( X \to Y \to Z \) of 5 moles of an ideal monoatomic gas, considering only P–V work.

Step 1: Given data and observations
- Number of moles, \( n = 5 \)
- Molar heat capacity at constant volume, \( C_{v,m} = 12 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Gas constant, \( R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Initial state \( X = (T=335\,K, V=10\,L) \)
- Intermediate state \( Y = (T=335\,K, V=20\,L) \)
- Final state \( Z = (T=415\,K, V=20\,L) \)

Step 2: Calculate \(\Delta H\) for each step
Enthalpy change for an ideal gas:
\[ \Delta H = n C_{p,m} \Delta T \] where \( C_{p,m} = C_{v,m} + R = 12 + 8.3 = 20.3 \, \text{J K}^{-1} \text{mol}^{-1} \).

Step 3: Step \( X \to Y \)
- Temperature constant (\(\Delta T = 0\)), so
\[ \Delta H_{X \to Y} = 5 \times 20.3 \times 0 = 0 \, \text{J} \]

Step 4: Step \( Y \to Z \)
- Temperature changes from 335 K to 415 K, so \(\Delta T = 80 \, K\)
- Volume constant (isochoric), so no P-V work but enthalpy changes with temperature
\[ \Delta H_{Y \to Z} = 5 \times 20.3 \times 80 = 8120 \, \text{J} \]

Step 5: Total enthalpy change \( X \to Y \to Z \)
\[ \Delta H = \Delta H_{X \to Y} + \Delta H_{Y \to Z} = 0 + 8120 = 8120 \, \text{J} \]

Final Answer:
\[ \boxed{8120 \, \text{J}} \]