Question

If a gas is heated at constant temperature, its isothermal compressibility
(A) Remains constant
(B) Increases linearly with temperature
(C) Decreases linearly with temperature
(D) Decreases inversely with temperature

Answer 1

Hint : Isothermal compressibility signifies the compressibility at constant temperature. The nature compressibility is simply gotten from its coefficient. We need to investigate the coefficient of compressibility for a temperature term.

Formula used: In this solution we will be using the following formula;
  $ \Rightarrow \alpha = - \dfrac{1}{V}{\left( {\dfrac{{\partial V}}{{\partial p}}} \right)_T} $ where $ \alpha $ is the coefficient of compressibility at constant temperature, $ V $ is the volume, and $ p $ is the pressure of the gas, and $ {\left( {\dfrac{{\partial V}}{{\partial p}}} \right)_T} $ signifies the volume change with pressure at constant temperature.

Complete step by step answer
The above gas in the question is said to be heated at constant temperature. When an object is heated at a constant temperature, this means that although it is heated, the pressure and volume vary with the thermal energy such that temperature is kept constant. In this case, the gas will certainly be allowed to do work to dissipate the internal energy due to the heat supplied.
The compressibility coefficient at constant temperature is given by,
 $ \Rightarrow \alpha = - \dfrac{1}{V}{\left( {\dfrac{{\partial V}}{{\partial p}}} \right)_T} $ where $ \alpha $ is the coefficient of compressibility at constant temperature, $ V $ is the volume, and $ p $ is the pressure of the gas, and $ {\left( {\dfrac{{\partial V}}{{\partial p}}} \right)_T} $ signifies the volume change with pressure at constant temperature. The negative sign counterbalances the fact that the volume of a substance decreases with increase in pressure, so as to make the compressibility coefficient a positive number.
Now, based on the partial differential equation above, no temperature term is present, hence the particular temperature has no effect on the compressibility of the gas at constant temperature.
Hence, the correct answer is A.

Note
In actuality, the general equation for volume change is given as
  $ \Rightarrow \dfrac{{dV}}{V} = \dfrac{1}{V}{\left( {\dfrac{{\partial V}}{{\partial p}}} \right)_T}dp + \dfrac{1}{V}{\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_p}dT $
  $ \Rightarrow \dfrac{{dV}}{V} = - \alpha dp + \beta dT $
The term on the far right is the expansion of the volume and $ \beta $ is the expansion coefficient of the gas.
We see that when $ dT = 0 $ , every temperature term drops out, and change in volume is independent on temperature.