Question

One mole of a monatomic ideal gas is expanded by a process described by \( PV^3 = C \), where \( C \) is a constant. The heat capacity of the gas during the process is given by (R is the gas constant)

• A. 2R
• B. 2.5R
• C. 1.5R
• D. R

Hint:

For a monatomic ideal gas, the heat capacity during a polytropic process can be derived using the equation \( C = \frac{3R}{2} \), considering the value of \( \gamma \).

Answer 1

The Correct Option is C

Step 1: A monatomic ideal gas molecule has 3 translational degrees of freedom and no rotational or vibrational contribution at ordinary temperatures.
Step 2: By the equipartition theorem, each degree of freedom contributes \( \tfrac{1}{2}R \) to the molar heat capacity, so the gas's heat capacity is \[ C = 3 \times \frac{1}{2}R = \frac{3}{2}R \]
Step 3: \[ \boxed{C = 1.5R} \]

Answer 2

A third way to frame this is by counting degrees of freedom directly and matching that count to each option, rather than working through \( C_v \)/\( C_p \) labels.

1. 2R: By the equipartition rule \( C = \tfrac{f}{2}R \), this value would correspond to \( f = 4 \) degrees of freedom, more than a monatomic gas possesses.
2. 2.5R: This corresponds to \( f = 5 \) degrees of freedom (translational plus 2 rotational), the situation for a diatomic gas, or equivalently to \( C_p \) of a monatomic gas; either way, not what a monatomic gas's own heat capacity represents.
3. 1.5R: This corresponds exactly to \( f = 3 \) degrees of freedom, translational motion only, which is precisely what a monatomic gas molecule has.
4. R: This corresponds to \( f = 2 \), fewer degrees of freedom than a real monatomic gas actually possesses.

Counting only the 3 translational degrees of freedom that a monatomic gas molecule genuinely has points to the \( \tfrac{3}{2}R \) value among the options.

Therefore, the correct answer is 1.5R.