Question

A vertical frictionless piston-cylinder arrangement contains air of mass 1 kg. During a process, 50 J of heat is transferred from outside to the system such that the piston is raised slowly by 0.1 m from its initial equilibrium position. The mass of the piston is 1 kg, and the diameter is 0.1 m. Assume that g = 9.81 m/s², and Patm = 100 kPa. The change in internal energy of the air in J (round off to two decimal places) lies between
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• A. 28.45 and 28.55
• B. -28.55 and -28.45
• C. -29.55 and -29.45
• D. 129.45 and 129.55

Hint:

Always be careful with the sign conventions for heat and work in thermodynamics. A common convention is: Heat added to the system is positive (+Q), and work done by the system is positive (+W). In the formula \(\Delta U = Q - W\), W is work done by the system.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires the application of the First Law of Thermodynamics to a closed system (the air inside the cylinder). The law states that the change in the internal energy of a system (\(\Delta U\)) is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)).
Step 2: Key Formula or Approach:
First Law of Thermodynamics: \(\Delta U = Q - W\)
Since the piston moves slowly against a constant external force (atmospheric pressure + piston weight), the process is a constant pressure (isobaric) expansion. The work done by the system is given by:
\(W = P \Delta V\), where P is the constant pressure of the air.
Step 3: Detailed Explanation or Calculation:
The system is the 1 kg of air inside the cylinder.
Heat Transfer (Q):
Heat is transferred *to* the system. So, Q is positive. \[ Q = +50 \text{ J} \] Work Done by the System (W):
The air expands and does work on its surroundings (the piston and the atmosphere).
First, calculate the constant pressure \(P\) of the air. The pressure balances the atmospheric pressure and the pressure exerted by the piston's weight. \[ P = P_{atm} + P_{piston} = P_{atm} + \frac{m_{piston} \times g}{A} \] Piston Area, \( A = \frac{\pi D^2}{4} = \frac{\pi (0.1)^2}{4} = \frac{0.01\pi}{4} \approx 0.007854 \text{ m}^2 \).
Pressure, \( P = 100 \times 10^3 \text{ Pa} + \frac{1 \text{ kg} \times 9.81 \text{ m/s}^2}{0.007854 \text{ m}^2} \approx 100000 + 1249.0 \text{ Pa} = 101249 \text{ Pa} \).
Next, calculate the change in volume \(\Delta V\). The piston is raised by \(h = 0.1\) m. \[ \Delta V = A \times h = 0.007854 \text{ m}^2 \times 0.1 \text{ m} = 0.0007854 \text{ m}^3 \] Now, calculate the work done \(W\): \[ W = P \times \Delta V = 101249 \text{ Pa} \times 0.0007854 \text{ m}^3 \approx 79.52 \text{ J} \] Change in Internal Energy (\(\Delta U\)):
Apply the First Law of Thermodynamics: \[ \Delta U = Q - W = 50 \text{ J} - 79.52 \text{ J} = -29.52 \text{ J} \] Step 4: Final Answer:
The change in internal energy of the air is -29.52 J.
Step 5: Why This is Correct:
The value -29.52 J lies in the range between -29.55 J and -29.45 J. Therefore, option (C) is the correct answer. The internal energy decreased because the air did more work on the surroundings than the heat it received.