Solution 17
We analyze the Pressure ($p$) vs Temperature ($T$) graph to deduce the movement of the pistons A and B.
Step 1: Analyzing the Graph Points
The graph shows three distinct phases:
- 240 K to 300 K: Line passes through origin (extrapolated). $P$ is proportional to $T$. This is an Isochoric Process (Constant Volume).
Pressure rises from $2.0 \times 10^5$ Pa to $2.5 \times 10^5$ Pa.
Since pressure increases without volume change, the pistons must be constrained. Initially, Piston A is at its maximum height $h_{max}$ (hitting the protrusion) while Piston B is still at the bottom. - 300 K to 450 K: Pressure is constant at $2.5 \times 10^5$ Pa. This is an Isobaric Process.
Volume must be increasing. This corresponds to Piston B rising from the bottom until it hits its protrusion at $h_{max}$. - 450 K to 540 K: Pressure rises again. Isochoric process. Both pistons are now at $h_{max}$.
Step 2: Calculating Areas ($A_A, A_B$)
At $T = 240$ K and $P = 2.0 \times 10^5$ Pa, only Piston A is filled (Piston B is at bottom).
$$V_1 = A_A h_{max} = A_A (1.0\,\text{m})$$
Using Ideal Gas Law ($PV = nRT$) for $n=1$ mole:
$$(2.0 \times 10^5)(A_A) = 1 \times 8.314 \times 240$$
$$A_A = \frac{1995.36}{2.0 \times 10^5} \approx 0.01\,\text{m}^2$$
At $T = 450$ K and $P = 2.5 \times 10^5$ Pa, both pistons are at $h_{max}$. $$V_2 = (A_A + A_B) h_{max} = A_A + A_B$$ Using the isobaric relation between 300K and 450K: $$\frac{V_{300}}{T_{300}} = \frac{V_{450}}{T_{450}} \implies \frac{A_A}{300} = \frac{A_A + A_B}{450}$$ $$1.5 A_A = A_A + A_B \implies A_B = 0.5 A_A$$ $$A_B = 0.005\,\text{m}^2$$
Step 3: Calculating Masses ($m_A, m_B$)
The pressure required to lift a piston is $P_{gas} = P_{atm} + \frac{mg}{S}$.
For Piston A: It supports the pressure at the start ($2.0 \times 10^5$ Pa). $$2.0 \times 10^5 = 1.0 \times 10^5 + \frac{m_A (10)}{0.01}$$ $$1.0 \times 10^5 = \frac{10 m_A}{0.01} \implies m_A = 100\,\text{kg}$$
For Piston B: It starts rising when pressure exceeds $2.5 \times 10^5$ Pa. $$2.5 \times 10^5 = 1.0 \times 10^5 + \frac{m_B (10)}{0.005}$$ $$1.5 \times 10^5 = \frac{10 m_B}{0.005} \implies m_B = 75\,\text{kg}$$
Summary of Results:
- Mass of A: 100 kg, Area of A: 0.01 m²
- Mass of B: 75 kg, Area of B: 0.005 m²