Question

in a long fixed horizontal pipe, 2 identical pistons a and b each of mass 500g can slide without friction. the pipe and te piston are made of perfectly insulating material. the space between pistons is filled with one mole helium. piston a is given initial velocity 10m/s twoards B. find max temp change of gas

Answer 1

1 K

Answer 2

Solution:

1. Total initial kinetic energy:
   Only piston A moves, so its energy is

   $$K_A=\frac{1}{2}\,m\,v^2=\frac{1}{2}(0.5)(10^2)=25\text{ J}.$$

2. Center‐of-mass (CM) motion:
   Initial CM velocity:

   $$v_{CM}=\frac{0.5\times10+0.5\times0}{1}=5\text{ m/s}.$$

   The kinetic energy associated with CM motion is

   $$K_{CM}=\frac{1}{2}(1)(5^2)=12.5\text{ J}.$$

3. Relative kinetic energy available for gas work:

   $$K_{\text{rel}}=25-12.5=12.5\text{ J}.$$

4. Conversion to gas's internal energy:
   Since the gas (1 mole of He, a monatomic gas) has internal energy

   $$U=\frac{3}{2}RT,$$

   the maximum temperature rise occurs when all $K_{\text{rel}}$ is converted, i.e.,

   $$\frac{3}{2}R\,\Delta T=12.5\text{ J}\quad\Rightarrow\quad \Delta T=\frac{2\times12.5}{3R}.$$

   Using $R = 8.314\, \text{J/mol·K}$,

   $$\Delta T=\frac{25}{3\times8.314}\approx\frac{25}{24.942}\approx 1\, \text{K}.$$

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Explanation (Minimal Core):

• Compute kinetic energy of piston A: $25\,J$.
• Find CM energy: $12.5\,J$ so that relative energy is $12.5\,J$.
• Set $\frac{3}{2}R\Delta T=12.5\,J$ and solve to get $\Delta T\approx1\,K$.