Question

1 mole of ideal diatomic gas is enclosed in a cylinder piston arrangement having cross-sectional area of piston \(4\,\text{cm}^2\). If gas has only rotational modes and \(P_{atm}=100\ \text{kPa}\), some amount of heat is added to the system as a result piston moves up slowly by \(2.5\,\text{cm}\). If temperature change is \(1.2^\circ C\). Find heat given to gas.

• A. \(19.9\,J\)
• B. \(23.5\,J\)
• C. \(14.6\,J\)
• D. \(10\,J\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The piston moves "slowly" against atmospheric pressure, indicating that the process is isobaric (constant pressure).
The gas has "only rotational modes" specified, which changes the degrees of freedom calculation.
Step 2: Key Formula or Approach:
Heat added in an isobaric process is $Q = n C_P \Delta T$.
Molar heat capacity at constant pressure is $C_P = C_V + R = \frac{f}{2}R + R$.
Step 3: Detailed Explanation:
Given $n = 1$ mole, $\Delta T = 1.2^\circ\text{C} = 1.2 \text{ K}$.
The gas has only rotational degrees of freedom, meaning $f = 2$.
Therefore, $C_V = \frac{2}{2}R = R$.
Consequently, $C_P = C_V + R = R + R = 2R$.
Now, calculate the heat given to the gas:
\[ Q = n C_P \Delta T \]
\[ Q = 1 \times (2R) \times 1.2 \]
\[ Q = 2.4 \times 8.314 \text{ J} \]
\[ Q = 19.9536 \text{ J} \approx 19.9 \text{ J} \]
Step 4: Final Answer:
The heat given to the gas is 19.9 J.