The volume V of a given mass of monoatomic gas changes with temperature T according to the relation `V=KT^(2//3)`. The workdone when temperature changes by 90 K will be xR. The value of x is [R = universal gas constant]

Answer

To solve the problem, we start with the given relationship between the volume \( V \) of a monoatomic gas and its temperature \( T \):

\[ V = k T^{\frac{2}{3}} \]

where \( k \) is a constant.

### Step 1: Relate Pressure and Volume
Using the ideal gas law, we have:

\[ PV = nRT \]

Assuming \( n = 1 \) (for simplicity), we can rewrite this as:

\[ PV = RT \]

From this, we can express \( T \) in terms of \( P \) and \( V \):

\[ T = \frac{PV}{R} \]

### Step 2: Substitute \( T \) into the Volume Equation
Now, substitute \( T \) into the volume equation:

\[ V = k \left(\frac{PV}{R}\right)^{\frac{2}{3}} \]

### Step 3: Simplify the Expression
This gives us:

\[ V = k \frac{(PV)^{\frac{2}{3}}}{R^{\frac{2}{3}}} \]

Rearranging, we have:

\[ V^{\frac{1}{3}} = \frac{k}{R^{\frac{2}{3}}} P^{\frac{2}{3}} \]

Let \( C = \frac{k}{R^{\frac{2}{3}}} \), then:

\[ PV^{\frac{1}{3}} = C \]

This indicates a polytropic process where \( PV^X = C \) with \( X = -\frac{1}{3} \).

### Step 4: Work Done in a Polytropic Process
The work done \( W \) in a polytropic process is given by:

\[ W = \frac{nR\Delta T}{1 - X} \]

Substituting \( n = 1 \), \( \Delta T = 90 \, K \), and \( X = -\frac{1}{3} \):

\[ W = \frac{R \cdot 90}{1 - (-\frac{1}{3})} = \frac{90R}{1 + \frac{1}{3}} = \frac{90R}{\frac{4}{3}} = \frac{90 \cdot 3R}{4} = \frac{270R}{4} = 67.5R \]

### Step 5: Compare with Given Format
Since the work done is expressed as \( xR \), we have:

\[ xR = 67.5R \]

Thus, comparing both sides gives:

\[ x = 67.5 \]

### Conclusion
The value of \( x \) is:

\[ \boxed{67.5} \]