One mole of an ideal monatomic gas undergoes the process `P=alphaT^(1//2)`, where `alpha` is constant . If molar heat capacity of the gas is `betaR1` when `R` = gas constant then find the value of `beta`.

Explanation

To solve the problem, we need to find the value of \(\beta\) for the given process \(P = \alpha T^{1/2}\) where \(\alpha\) is a constant, and the molar heat capacity is given as \(\beta R\).

### Step 1: Understand the relationship between P, V, and T
For an ideal gas, the equation of state is given by:
\[ PV = nRT \]
Since we have one mole of gas (\(n = 1\)), we can write:
\[ PV = RT \]

### Step 2: Substitute the expression for P
From the given process, we have:
\[ P = \alpha T^{1/2} \]
Substituting this into the ideal gas equation:
\[ \alpha T^{1/2} V = RT \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ V = \frac{RT}{\alpha T^{1/2}} = \frac{R}{\alpha} T^{1/2} \]

### Step 4: Differentiate to find dQ/dT
We need to find the heat capacity \(C\). The heat added \(dQ\) can be expressed as:
\[ dQ = dU + dW \]
For a monatomic ideal gas, the internal energy \(U\) is given by:
\[ dU = C_v dT = \frac{3}{2} R dT \]
The work done \(dW\) is given by:
\[ dW = P dV \]
Substituting \(P\) and \(dV\):
\[ dW = \alpha T^{1/2} dV \]

### Step 5: Find dV
Differentiating \(V\) with respect to \(T\):
\[ dV = \frac{R}{\alpha} \cdot \frac{1}{2} T^{-1/2} dT = \frac{R}{2\alpha} T^{-1/2} dT \]

### Step 6: Substitute dV in dW
Substituting \(dV\) into \(dW\):
\[ dW = \alpha T^{1/2} \cdot \frac{R}{2\alpha} T^{-1/2} dT = \frac{R}{2} dT \]

### Step 7: Find dQ
Now substituting \(dU\) and \(dW\) into \(dQ\):
\[ dQ = dU + dW = \frac{3}{2} R dT + \frac{R}{2} dT = 2R dT \]

### Step 8: Find the molar heat capacity C
The molar heat capacity \(C\) is defined as:
\[ C = \frac{dQ}{dT} = 2R \]

### Step 9: Relate C to \(\beta\)
Given that the molar heat capacity is also expressed as \(\beta R\):
\[ \beta R = 2R \]
Thus, we can conclude:
\[ \beta = 2 \]

### Final Answer
The value of \(\beta\) is \(2\).

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