In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $$\Delta T$$. The amount of heat absorbed by gas is (R is gas constant):

We have one mole of an ideal mono-atomic gas. Throughout the process the variables obey $$VT = K$$ where $$K$$ is a positive constant. Because the gas is ideal, the equation of state is also valid at every instant:

$$PV = RT \qquad\text{(for }n = 1\text{ mol).}$$

The temperature is raised by $$\Delta T,$$ so if we call the initial temperature $$T_i,$$ the final temperature is $$T_f = T_i + \Delta T.$$

First we express the volume in terms of temperature using the given relation:

$$VT = K \;\;\Longrightarrow\;\; V = \dfrac{K}{T}.$$

Next we write the pressure in terms of temperature alone. Substituting this $$V$$ into the ideal-gas equation gives

$$P = \dfrac{RT}{V} = \dfrac{RT}{K/T} = \dfrac{RT^{2}}{K}.$$

To find the work done by the gas we use the formula

$$W = \int_{V_i}^{V_f} P\,dV.$$

Because it is more convenient to integrate over temperature, we change variables. Differentiate the relation $$V = K/T$$:

$$dV = -\,\dfrac{K}{T^{2}}\;dT.$$

Now substitute both $$P$$ and $$dV$$ into the integral:

$$\displaystyle W = \int_{T_i}^{T_f} \Bigl(\dfrac{RT^{2}}{K}\Bigr)\,\Bigl(-\,\dfrac{K}{T^{2}}\;dT\Bigr) = \int_{T_i}^{T_f} \bigl(-R\bigr)\,dT = -R\,(T_f - T_i).$$

Since $$T_f - T_i = \Delta T,$$ the work done by the gas is

$$W = -\,R\Delta T.$$

The negative sign means the gas has been compressed (its volume decreased as temperature rose), so work is done on the gas; nevertheless, we keep the sign in further calculations exactly as obtained.

For a mono-atomic ideal gas the molar heat capacity at constant volume is

$$C_V = \dfrac{3}{2}R.$$

The change in internal energy is therefore, by the formula $$\Delta U = nC_V\Delta T,$$

$$\Delta U = (1)\Bigl(\dfrac{3}{2}R\Bigr)\Delta T = \dfrac{3}{2}R\Delta T.$$

Finally, we apply the first law of thermodynamics, $$Q = \Delta U + W.$$ Substituting the values just found gives

$$Q = \dfrac{3}{2}R\Delta T + \bigl(-R\Delta T\bigr) = \Bigl(\dfrac{3}{2} - 1\Bigr)R\Delta T = \dfrac{1}{2}R\Delta T.$$

This $$Q$$ is positive, so that amount of heat is indeed absorbed by the gas.

Hence, the correct answer is Option A.