For a fixed amouunt of an ideal gas`(gamma=(11)/(9))`, the change in internal energy of the gas when pressure changes from 10 bar to 20 bar in rigid vessel of volume 5 L is given by :

Options

A. 225 J
B. 22.5 kJ
C. 15 kJ
D. 36 kJ

Answer

To solve the problem of finding the change in internal energy of an ideal gas when the pressure changes from 10 bar to 20 bar in a rigid vessel of volume 5 L, we can follow these steps:

### Step 1: Understand the relationship between internal energy, temperature, and specific heat
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
Where:
- \( n \) = number of moles of the gas
- \( C_V \) = molar specific heat at constant volume
- \( \Delta T \) = change in temperature

### Step 2: Relate \( C_V \) and \( C_P \) using the given gamma (γ)
We are given that:
\[ \gamma = \frac{C_P}{C_V} = \frac{11}{9} \]
From the relationship between \( C_P \) and \( C_V \):
\[ C_P = C_V + R \]
We can express \( C_V \) in terms of \( R \):
\[ \frac{C_P}{C_V} = \frac{11}{9} \implies C_P = \frac{11}{9} C_V \]
Substituting \( C_P \) in the equation:
\[ \frac{11}{9} C_V = C_V + R \]
Rearranging gives:
\[ \frac{11}{9} C_V - C_V = R \implies \frac{2}{9} C_V = R \implies C_V = \frac{9}{2} R \]

### Step 3: Calculate the change in temperature (ΔT)
Using the ideal gas law:
\[ PV = nRT \implies T = \frac{PV}{nR} \]
We need to find the change in temperature as pressure changes from 10 bar to 20 bar while volume remains constant (5 L = 0.005 m³).

1. Calculate initial temperature \( T_1 \):
\[ T_1 = \frac{P_1 V}{nR} = \frac{10 \times 10^5 \times 0.005}{nR} \]
(Convert bar to pascal: 1 bar = \( 10^5 \) Pa)

2. Calculate final temperature \( T_2 \):
\[ T_2 = \frac{P_2 V}{nR} = \frac{20 \times 10^5 \times 0.005}{nR} \]

3. The change in temperature \( \Delta T \):
\[ \Delta T = T_2 - T_1 = \frac{(20 - 10) \times 10^5 \times 0.005}{nR} = \frac{10 \times 10^5 \times 0.005}{nR} \]

### Step 4: Substitute values into the internal energy equation
Now substituting \( C_V \) and \( \Delta T \) into the internal energy equation:
\[ \Delta U = n C_V \Delta T = n \left(\frac{9}{2} R\right) \left(\frac{10 \times 10^5 \times 0.005}{nR}\right) \]
The \( n \) and \( R \) cancel out:
\[ \Delta U = \frac{9}{2} \times 10 \times 10^5 \times 0.005 \]
Calculating this gives:
\[ \Delta U = \frac{9 \times 10 \times 10^5 \times 0.005}{2} = \frac{45000}{2} = 22500 \text{ J} = 22.5 \text{ kJ} \]

### Final Answer
The change in internal energy of the gas is:
\[ \Delta U = 22.5 \text{ kJ} \]