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. \(10\,J\)

Hint:

Split the heat added using the first law, $Q=\Delta U + W$. Find the internal energy change from the degrees of freedom given (only rotational, so $f=2$), and find the work done directly from $P\Delta V = nR\Delta T$ rather than the piston dimensions.

Answer 1

The Correct Option is A

Concept: Since the piston moves slowly, the process is isobaric (pressure constant). Heat supplied: \[ Q = nC_p\Delta T \] For a diatomic gas with only rotational modes: \[ f = 2 \] \[ C_p = \left(\frac{f}{2}+1\right)R \]
Step 1:Find molar heat capacity at constant pressure. \[ C_p = \left(\frac{2}{2}+1\right)R \] \[ C_p = 2R \]
Step 2:Calculate heat supplied. \[ Q = nC_p\Delta T \] \[ Q = 1 \times 2R \times 1.2 \] \[ Q = 2 \times 8.314 \times 1.2 \] \[ Q = 19.95\,J \] \[ \boxed{Q \approx 19.9\,J} \]

Answer 2

Concept:

• The first law of thermodynamics splits heat added into two parts, the rise in internal energy and the work done by the gas, $Q = \Delta U + W$.
• For an ideal gas at constant pressure, the work done during expansion can be written directly in terms of temperature change using $PV=nRT$, without needing the volume change itself.
• With only rotational modes active, the gas has $f=2$ degrees of freedom, so $C_V = R$.

Step 1: Identify the type of process.
The piston moves slowly against a constant atmospheric pressure, so this is a quasi-static, constant pressure (isobaric) process.

Step 2: Find the internal energy change.
With only rotational modes active, $f=2$, so $C_V = \dfrac{f}{2}R = R$.
$\Delta U = nC_V\Delta T = 1 \times R \times 1.2$
$\Delta U = 1 \times 8.314 \times 1.2 = 9.98\ J$

Step 3: Find the work done by the gas without using the volume change directly.
At constant pressure, $P\Delta V = nR\Delta T$ directly from the ideal gas equation, so $W = nR\Delta T$.
$W = 1 \times 8.314 \times 1.2 = 9.98\ J$

Step 4: Apply the first law of thermodynamics.
$Q = \Delta U + W$
$Q = 9.98 + 9.98$
$Q = 19.95\ J$

Final Answer: $Q \approx 19.9\ J$