One mole of an ideal monoatomic gas is taken from state A to state B through the process `P=3/2 T^(1//2)` It is found that its temperature increases by 100 K in this process. Now it is taken from state B to C through a process for which internal energy is related to volume as `U= 1/2 V^(1//2)` Find the total work performed by the gas (in Joule) if it is given that volume at B is `100 m^(3)` and at C it is `1600m^(3)` [Use `R= 8.3 J //mol-K`]

Explanation

To solve the problem, we need to calculate the total work done by the gas during the two processes: from state A to state B and from state B to state C.

### Step 1: Work Done from A to B

The process from state A to state B is defined by the equation \( P = \frac{3}{2} T^{1/2} \). We know that for an ideal gas, the pressure \( P \) can also be expressed as:

\[ P = nRT \]

where \( n \) is the number of moles and \( R \) is the universal gas constant. Given that \( n = 1 \) mole and \( R = 8.3 \, \text{J/mol-K} \), we can equate the two expressions for pressure:

\[ \frac{3}{2} T^{1/2} = R T \]

### Step 2: Relate Volume and Temperature

From the ideal gas law, we can express volume \( V \) in terms of temperature \( T \):

\[ PV = nRT \implies V = \frac{nRT}{P} \]

Substituting \( P \):

\[ V = \frac{nRT}{\frac{3}{2} T^{1/2}} = \frac{2nR}{3} T^{1/2} \]

### Step 3: Differentiate Volume with Respect to Temperature

To find the work done, we need to express \( dV \) in terms of \( dT \):

Differentiating \( V \):

\[ dV = \frac{2nR}{3} \cdot \frac{1}{2} T^{-1/2} dT = \frac{nR}{3} T^{-1/2} dT \]

### Step 4: Substitute into Work Integral

The work done \( W_{AB} \) from A to B is given by:

\[ W_{AB} = \int P \, dV \]

Substituting \( P \) and \( dV \):

\[ W_{AB} = \int \frac{3}{2} T^{1/2} \cdot \frac{nR}{3} T^{-1/2} dT = \int \frac{nR}{2} dT \]

### Step 5: Evaluate the Integral

The temperature increases by \( 100 \, K \), so we can integrate from \( T_A \) to \( T_B \):

\[ W_{AB} = \frac{nR}{2} \int_{T_A}^{T_B} dT = \frac{nR}{2} (T_B - T_A) = \frac{1 \cdot 8.3}{2} \cdot 100 = 415 \, J \]

### Step 6: Work Done from B to C

Now, we need to find the work done from state B to state C, where the internal energy \( U \) is related to volume as:

\[ U = \frac{1}{2} V^{1/2} \]

The work done \( W_{BC} \) can be calculated using the relationship between internal energy and work:

\[ W_{BC} = \Delta U = U_C - U_B \]

### Step 7: Calculate Internal Energies

At volume \( V_B = 100 \, m^3 \):

\[ U_B = \frac{1}{2} (100)^{1/2} = \frac{1}{2} \cdot 10 = 5 \, J \]

At volume \( V_C = 1600 \, m^3 \):

\[ U_C = \frac{1}{2} (1600)^{1/2} = \frac{1}{2} \cdot 40 = 20 \, J \]

### Step 8: Calculate Work Done from B to C

Thus, the work done from B to C is:

\[ W_{BC} = U_C - U_B = 20 \, J - 5 \, J = 15 \, J \]

### Step 9: Total Work Done

The total work done by the gas is:

\[ W_{total} = W_{AB} + W_{BC} = 415 \, J + 15 \, J = 430 \, J \]

### Final Answer

The total work performed by the gas is:

\[ \boxed{430 \, J} \]