One mole of an ideal monatomic gas undergoes the process `p=alphaT^(1//2)`, where `alpha` is a constant.
(a) Find the work done by the gas if its temperature increases by 50K.
(b) Also, find the molar specific heat of the gas.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Finding the Work Done by the Gas

1. **Identify the Process**: The process is given by the equation \( p = \alpha T^{1/2} \). We can rewrite this in a form that relates pressure and volume. From the ideal gas law \( PV = nRT \), we can express \( T \) in terms of \( p \) and \( V \):
\[ T = \frac{pV}{nR} \]

2. **Express the Relationship**: We can manipulate the given equation:
\[ p = \alpha T^{1/2} \implies pT^{-1/2} = \alpha \]
This indicates that \( pT^{-1/2} \) is a constant.

3. **Identify the Polytropic Process**: The relationship \( pV^x = \text{constant} \) can be compared with our derived equation. By rearranging, we find:
\[ p^{1/2}V^{-1/2} = \text{constant} \]
This implies \( x = -1 \).

4. **Calculate Work Done**: For a polytropic process, the work done \( W \) is given by:
\[ W = \frac{nR\Delta T}{1 - x} \]
Here, \( n = 1 \) mole, \( R = 8.31 \, \text{J/(mol K)} \), \( \Delta T = 50 \, \text{K} \), and \( x = -1 \):
\[ W = \frac{1 \cdot 8.31 \cdot 50}{1 - (-1)} = \frac{415.5}{2} = 207.75 \, \text{J} \]

### Part (b): Finding the Molar Specific Heat of the Gas

1. **Use the Formula for Molar Specific Heat**: The molar specific heat \( C \) for a polytropic process is given by:
\[ C = C_v + \frac{R}{1 - x} \]
For a monatomic ideal gas, the molar specific heat at constant volume \( C_v = \frac{3}{2}R \).

2. **Substitute Values**: Now substituting \( C_v \) and \( x \):
\[ C = \frac{3}{2}R + \frac{R}{1 - (-1)} = \frac{3}{2}R + \frac{R}{2} = \frac{3R + R}{2} = \frac{4R}{2} = 2R \]

### Final Answers
- (a) The work done by the gas is \( 207.75 \, \text{J} \).
- (b) The molar specific heat of the gas is \( 2R \).