Question

A cylinder closed at both ends is separated into two equal parts (45 cm each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 K and a pressure of 1 atm. How much the gas should be heated in one part of the cylinder to shift the piston by 5 cm and the pressure of the gas after shifting the piston?

• A. $T = 365$ K and $P = 1.125$ atm
• B. $T = 350$ K and $P = 1.125$ atm
• C. $T = 375$ K and $P = 2.125$ atm
• D. $T = 350$ K and $P = 2.125$ atm
• E. $T = 375$ K and $P = 1.125$ atm

Hint:

Always start with the side that maintains a constant temperature. Its pressure change is purely determined by the volume change ($P \propto 1/V$), which then gives you the final pressure for the whole system.

Answer 1

Answer: (E)

Concept:
For an ideal gas, we use the equation $PV = nRT$. Since the mass (and $n$) is the same in both parts, we use the relationship: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] When the piston shifts and stabilizes, the pressure on both sides must be equal ($P_{final}$).

Step 1: Analyze the part that is NOT heated.
Let Part A be heated and Part B remain at 300 K. Initially: $V_B = 45A$ (where $A$ is cross-section), $P_B = 1$ atm, $T_B = 300$ K. Finally: The piston moves 5 cm away from Part A into Part B. $V_B' = (45 - 5)A = 40A$. $T_B$ remains 300 K. \[ P_{final} = \frac{P_B V_B}{V_B'} = \frac{1 \times 45A}{40A} = 1.125 \text{ atm} \]

Step 2: Analyze the part that IS heated (Part A).
Initially: $V_A = 45A, P_A = 1 \text{ atm}, T_A = 300$ K. Finally: $V_A' = (45 + 5)A = 50A, P_{final} = 1.125 \text{ atm}, T_{final} = ?$ \[ \frac{P_A V_A}{T_A} = \frac{P_{final} V_A'}{T_{final}} \implies \frac{1 \times 45}{300} = \frac{1.125 \times 50}{T_{final}} \] \[ T_{final} = \frac{1.125 \times 50 \times 300}{45} = \frac{56.25 \times 300}{45} = 1.25 \times 300 = 375 \text{ K} \]