Question

For a monoatomic gas, the work done at constant pressure is $W$. The heat supplied at constant volume for the same rise in temperature of the gas is

• A. $2W$
• B. $W$
• C. $\frac{W}{2}$
• D. $\frac{3W}{2}$

Hint:

For an ideal gas, work done at constant pressure is always $nR\Delta T$, while the internal energy change (which equals heat at constant volume) is $C_v$ scaled by $n\Delta T$. The ratio of $Q_v$ to $W$ is simply the ratio $\frac{C_v}{R}$. For a monoatomic gas, this ratio is automatically $\frac{3}{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem presents an ideal monoatomic gas undergoing two separate thermodynamic operations that share the exact same temperature elevation ($\Delta T$).
In the first path, under constant pressure ($P$), the gas performs mechanical work equal to $W$. We need to find the equivalent heat needed ($Q_v$) to produce that identical temperature jump under constant volume ($V$) constraints.

Step 2: Key Formula or Approach:
1. Isobaric work done at constant pressure is defined by:
$$W = P\Delta V = nR\Delta T$$ 2. Isochoric heat transfer at constant internal volume is defined by:
$$Q_v = nC_v\Delta T$$ 3. For a standard monoatomic ideal gas, the molar heat capacity at a fixed volume is defined by the degree of freedom relation:
$$C_v = \frac{3}{2}R$$

Step 3: Detailed Explanation:
From the work formula at constant pressure, we isolate the common temperature parameter group:
$$nR\Delta T = W$$ Now evaluate the required heat expression at constant volume:
$$Q_v = nC_v\Delta T$$ Substitute the specific monoatomic gas value $C_v = \frac{3}{2}R$ into this expression:
$$Q_v = n\left(\frac{3}{2}R\right)\Delta T = \frac{3}{2}(nR\Delta T)$$ We can directly substitute the initial isobaric work identity ($nR\Delta T = W$) into this relation:
$$Q_v = \frac{3}{2}W$$

Step 4: Final Answer:
The heat supplied at constant volume for the same rise in temperature is $\frac{3W}{2}$, matching option (D).