Question

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 incresed by $\Delta T$. The amount of heat absorbed by gas is (R is gas constant) :

• A. $\frac{1}{2} R \Delta T$
• B. $\frac{3}{2} R \Delta T$
• C. $\frac{1}{2} K R \Delta T$
• D. $\frac{2K}{3} \Delta T$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the heat absorbed by an ideal monoatomic gas undergoing a process where the product of its volume and temperature remains constant. The relation given is \( VT = K \), where \( K \) is a constant and the gas temperature increases by \( \Delta T \).

Let's start by analyzing the given relation \( VT = K \). Differentiating both sides with respect to time, we get:

\[ T \frac{dV}{dt} + V \frac{dT}{dt} = 0 \]

This implies a change in temperature \( \Delta T \) is associated with a proportional change in volume \( \Delta V \) such that the product remains constant. For this derivative to add up to zero, we find:

\[ T \Delta V = - V \Delta T \]

The work done by/on the gas during this process can be given by the expression:

\[ W = P \Delta V \]

Using the ideal gas law in its form \( PV = nRT \) (where \( n = 1 \) mole for this problem), we substitute \( P = \frac{nRT}{V} \) to get the work done:

\[ W = \frac{RT}{V} \Delta V \]

Using the relation derived from \( VT = K \), \( \Delta V = -\frac{V}{T} \Delta T \), we substitute in for work:

\[ W = \frac{RT}{V} \left(-\frac{V}{T} \Delta T\right) = -R \Delta T \]

In an ideal monoatomic gas, the change in internal energy (\( \Delta U \)) for one mole is given by:

\[ \Delta U = \frac{3}{2} R \Delta T \]

Using the first law of thermodynamics: \( \Delta U = Q - W \), where \( Q \) is the heat absorbed by the system, we now substitute the expressions for \( \Delta U \) and \( W \):

\[ \frac{3}{2} R \Delta T = Q + R \Delta T \]

Solving for \( Q \), we find:

\[ Q = \frac{3}{2} R \Delta T + R \Delta T \] \[ Q = \frac{1}{2} R \Delta T \]

Therefore, the correct option is:

\(\frac{1}{2} R \Delta T\)

This result shows us how heat absorption reflects changes in kinetic energy when conditions adjust isotropically. The other options presented don't match the derived formula under the outlined principles of gas behavior. Therefore, the solution to the problem with the condition given is option (a).