1. Given relation

For one mole of an ideal monoatomic gas, $VT=K$ where $K$ is constant.

Also, temperature increases by $\Delta T$.

We need the heat absorbed, $\delta Q$.

2. Use first law of thermodynamics

For an infinitesimal change, $\delta Q=dU+\delta W$

For one mole of a monoatomic ideal gas, $dU=nC_V dT=\frac{3}{2}R\,dT$ since $n=1$ and $C_V=\frac{3}{2}R$.

Work done by gas: $\delta W=P\,dV$

So, $\delta Q=\frac{3}{2}R\,dT+P\,dV$

3. Express $P\,dV$ using the given constraint

From ideal gas equation for one mole, $PV=RT$ so $P=\frac{RT}{V}$

Given, $VT=K \implies V=\frac{K}{T}$

Differentiate: $dV=-\frac{K}{T^2}dT$

Now, $P\,dV=\frac{RT}{V}dV$

Using $V=K/T$, $P=\frac{RT}{K/T}=\frac{RT^2}{K}$

Hence, $P\,dV=\frac{RT^2}{K}\left(-\frac{K}{T^2}dT\right)=-R\,dT$

4. Find heat absorbed

Therefore, $\delta Q=\frac{3}{2}R\,dT-R\,dT=\frac{1}{2}R\,dT$

Integrating for temperature increase $\Delta T$, $Q=\int \delta Q=\frac{1}{2}R\int dT=\frac{1}{2}R\Delta T$

5. Match with options

$Q=\frac{1}{2}R\Delta T$

So the correct option is:

B. $\frac{1}{2}R\Delta T$

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.