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Step-by-Step Solution
Step 1: Write down the known relations
• We have a monoatomic gas whose volume is given by the relation
V = K \, T^{\frac{2}{3}} .
• The work done W during an infinitesimal process is given by the integral
W = \int P\, dV .
Step 2: Express pressure in terms of temperature and volume
From the ideal gas law for n moles of the gas,
P = \frac{nRT}{V} .
Step 3: Substitute P in the work integral
Substitute P = \frac{nRT}{V} into W = \int P \, dV to get:
W = \int \frac{nRT}{V} \, dV.
Step 4: Relate dV and dT using V = K \, T^{\frac{2}{3}}
Given V = K \, T^{\frac{2}{3}} ,
we can differentiate both sides with respect to T :
\frac{dV}{dT} = \frac{2}{3} K \, T^{-\frac{1}{3}} \quad \Rightarrow \quad dV = \frac{2}{3} K \, T^{-\frac{1}{3}} \, dT.
Step 5: Rewrite the work integral in terms of T
In the expression for work,
W = \int \frac{nRT}{V} \, dV,
substitute V = K \, T^{\frac{2}{3}} and dV = \frac{2}{3} K \, T^{-\frac{1}{3}} \, dT :
W = \int \frac{nRT}{K\, T^{\frac{2}{3}}} \cdot \frac{2}{3}K \, T^{-\frac{1}{3}} \, dT.
The K terms cancel out, and combining powers of T gives:
W = \int \frac{nR \, 2}{3} \, T \, T^{-\frac{1}{3}} \, T^{-\frac{2}{3}} \, dT = \int \frac{2}{3} nR \, dT,
since T \times T^{-\frac{1}{3}} = T^{\frac{2}{3}} and dividing by T^{\frac{2}{3}} cancels everything except for a constant factor.
Thus,
W = \frac{2}{3} nR \int dT = \frac{2}{3} nR \, (T_2 - T_1).
Step 6: Apply the given temperature change
We are given that the temperature changes by 90\,\text{K} , so if T_2 - T_1 = 90 , then the work done is:
W = \frac{2}{3} nR \times 90 = 60 \, nR.
Step 7: Evaluate for one mole
Assuming n = 1 mole of gas,
W = 60 \, R.
Hence, the value of x in xR is 60.
Answer: 60
