Question

A monoatomic ideal gas is heated at constant pressure. The percentage of total heat used in increasing the internal energy and that used for doing external work is $A$ and $B$ respectively. Then the ratio, $\text{A} : \text{B}$ is

• A. $5 : 3$
• B. $2 : 3$
• C. $3 : 2$
• D. $2 : 5$

Hint:

For monoatomic gases at constant $P$, 60% of heat goes to internal energy and 40% to work.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When heat is added to a gas at constant pressure, part of it goes into increasing the internal energy (raising temperature) and the rest goes into doing external work (expanding volume), as per the First Law of Thermodynamics.
The ratio of these portions depends on the specific heat capacities of the gas, which in turn depend on its atomicity.
Step 2: Key Formula or Approach:
First Law of Thermodynamics: $Q = \Delta U + W$.
Heat added at constant pressure: $Q = n C_p \Delta T$.
Change in internal energy: $\Delta U = n C_v \Delta T$.
Work done: $W = P \Delta V = n R \Delta T$.
The ratio $\text{A} : \text{B}$ is the ratio of internal energy increase to work done: $\frac{A}{B} = \frac{\Delta U}{W}$.
Step 3: Detailed Explanation:
For a monoatomic ideal gas, the molar specific heat at constant volume is $C_v = \frac{3}{2} R$.
The portion of heat $A$ goes to internal energy $\Delta U$: \[ A \propto \Delta U = n C_v \Delta T = n \left(\frac{3}{2} R\right) \Delta T \] The portion of heat $B$ goes to external work $W$: \[ B \propto W = n R \Delta T \] We want to find the ratio $\text{A} : \text{B}$: \[ \text{Ratio} = \frac{\Delta U}{W} = \frac{n \left(\frac{3}{2} R\right) \Delta T}{n R \Delta T} \] Cancel the common terms $n$, $R$, and $\Delta T$: \[ \text{Ratio} = \frac{3/2}{1} = \frac{3}{2} \] Therefore, the ratio $\text{A} : \text{B}$ is $3 : 2$.
Step 4: Final Answer:
The ratio is $3 : 2$.