One mole of a real gas is subjected to a process from (2 bar , 30 lit ., 300k) to (2 bar , 40 , lit ., 500K)
Given :`C_(v)=25 J//"mole"/K`
`C_(p)=40 J//"mole"/K`
Calculate `DeltaU`.

Options

A. 5000 J
B. 6000 J
C. 8000 J
D. 10000 J

Explanation

To calculate the change in internal energy (ΔU) for one mole of a real gas undergoing a process from (2 bar, 30 L, 300 K) to (2 bar, 40 L, 500 K), we can follow these steps:

### Step 1: Calculate the change in temperature (ΔT)
Given:
- Initial temperature (T1) = 300 K
- Final temperature (T2) = 500 K

The change in temperature (ΔT) can be calculated as:
\[ \Delta T = T2 - T1 = 500 \, \text{K} - 300 \, \text{K} = 200 \, \text{K} \]

### Step 2: Calculate the heat added (Q)
Using the formula for heat added (Q) at constant pressure:
\[ Q = n \cdot C_p \cdot \Delta T \]
Where:
- n = number of moles = 1 mole
- \(C_p = 40 \, \text{J/mole/K}\)

Substituting the values:
\[ Q = 1 \cdot 40 \cdot 200 = 8000 \, \text{J} \]

### Step 3: Calculate the change in volume (ΔV)
Given:
- Initial volume (V1) = 30 L
- Final volume (V2) = 40 L

The change in volume (ΔV) can be calculated as:
\[ \Delta V = V2 - V1 = 40 \, \text{L} - 30 \, \text{L} = 10 \, \text{L} \]

### Step 4: Calculate the work done (W)
The work done (W) by the gas at constant pressure can be calculated using:
\[ W = -P_{\text{external}} \cdot \Delta V \]
Where:
- \(P_{\text{external}} = 2 \, \text{bar}\)
- To convert bar to Joules, we use the conversion factor \(1 \, \text{bar} = 100 \, \text{J/L}\).

Thus, we convert the pressure:
\[ W = -2 \, \text{bar} \cdot 10 \, \text{L} \cdot 100 \, \text{J/L} = -2000 \, \text{J} \]

### Step 5: Calculate the change in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = Q + W \]
Substituting the values we calculated:
\[ \Delta U = 8000 \, \text{J} + (-2000 \, \text{J}) = 6000 \, \text{J} \]

### Final Answer:
\[ \Delta U = 6000 \, \text{J} \]

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