Question

A sample of an ideal gas ($\gamma = 5/3$) is heated at constant pressure. If 100 J of heat is supplied to the gas, the work done by the gas is ______.

• A. 150 J
• B. 60 J
• C. 40 J
• D. 250 J

Hint:

Memorize the energy distribution ratios for isobaric heating of a monoatomic gas ($\gamma = 5/3$):
$\Delta U : W : Q \implies 3 : 2 : 5$.
If $Q = 100$, then $W = 40$ and $\Delta U = 60$. It solves the problem in 2 seconds!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An ideal gas undergoes an isobaric (constant pressure) expansion. We are given the total heat supplied ($Q$) and the ratio of specific heats ($\gamma$), and must calculate the mechanical work done ($W$) by the gas.

Step 2: Key Formula or Approach:
For an isobaric process:
1. Heat supplied: $Q = n C_p \Delta T$
2. Work done: $W = P \Delta V = n R \Delta T$
The ratio of work done to heat supplied is a highly useful standard relationship:
$$\frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p}$$
Using Meyer's relation ($C_p - C_v = R$) and $\gamma = C_p/C_v$, we know $C_p = \frac{\gamma R}{\gamma - 1}$.
Substituting this yields the fraction of heat converted to work:
$$\frac{W}{Q} = \frac{\gamma - 1}{\gamma} = 1 - \frac{1}{\gamma}$$

Step 3: Detailed Explanation:
Given:
Heat supplied $Q = 100 \text{ J}$
Gas constant ratio $\gamma = 5/3$ (indicating a monoatomic gas).
Use the derived ratio formula:
$$\frac{W}{Q} = \frac{\frac{5}{3} - 1}{\frac{5}{3}}$$
$$\frac{W}{Q} = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}$$
This means exactly 2/5 (or 40%) of the total heat supplied goes into doing external work, while the remaining 3/5 (60%) goes into raising internal energy.
Calculate the work done $W$:
$$W = \frac{2}{5} \times Q$$
$$W = \frac{2}{5} \times 100 \text{ J} = 2 \times 20 \text{ J} = 40 \text{ J}$$

Step 4: Final Answer:
The work done by the gas is 40 J, matching option (c).