Question

A cylindrical container is divided into three parts by two light fitting pistons. The pistons are connected by a spring. The region between the pistons is vacuum and the other two parts have same number of moles of an ideal gas initially, both the gas chambers are at temperature T0 and the spring is compressed by 1m. Length of both gas chambers is 1m in this position. Now the temperature of the left and right chambers are raised to $\frac{4T_0}{3}$ and $\frac{5T_0}{3}$ respectively. The final compression in the spring in equilibrium is given by $\frac{\sqrt{k-1}}{2}$.

Find k (All the walls and pistons are insulating)

Answer 1

13

Answer 2

The problem describes a cylindrical container divided into three parts by two light fitting pistons. The central region is a vacuum, and the two outer regions (left and right chambers) contain an ideal gas. A spring connects the two pistons.

1. Initial State Analysis:

• Let the cross-sectional area of the cylinder be A.
• Initial length of both gas chambers is 1m. So, $V_{L,0} = V_{R,0} = A \times 1 = A$.
• Both gas chambers have the same number of moles (n) and temperature ($T_0$).
• From the ideal gas law, $P_{L,0} V_{L,0} = nRT_0$ and $P_{R,0} V_{R,0} = nRT_0$.
• Since $V_{L,0} = V_{R,0}$ and $T_{L,0} = T_{R,0}$, the initial pressures must be equal: $P_{L,0} = P_{R,0} = P_0$.
• The spring is initially compressed by $x_0 = 1m$. Let the spring constant be K.
• For the system to be in equilibrium, the force exerted by the gas on each piston must balance the force exerted by the spring.
  • For the left piston: $P_0 A = Kx_0 = K(1)$.
  • For the right piston: $P_0 A = Kx_0 = K(1)$.
• From the ideal gas law for one chamber: $P_0 A = nRT_0 / 1m$.
• Equating the forces: $K(1) = nRT_0$. This gives us a relationship between K, n, R, and $T_0$.

2. Final State Analysis:

• The temperatures are changed: $T_L = \frac{4T_0}{3}$ and $T_R = \frac{5T_0}{3}$.
• Let the final compression in the spring be $x$.
• Let the final lengths of the gas chambers be $L_L$ and $L_R$.
• In the final equilibrium, the force balance requires:
  • For the left piston: $P_L A = Kx$.
  • For the right piston: $P_R A = Kx$.
• This implies $P_L = P_R = P_{final} = \frac{Kx}{A}$.

3. Applying Ideal Gas Law for Final State:

• For the left chamber: $P_L V_L = nRT_L \implies P_L (A L_L) = nR \left(\frac{4T_0}{3}\right)$.

  • Substitute $P_L = \frac{Kx}{A}$: $\left(\frac{Kx}{A}\right) (A L_L) = nR \left(\frac{4T_0}{3}\right)$.
  • $Kx L_L = nR \left(\frac{4T_0}{3}\right)$.
  • From the initial state, we found $K = nRT_0$. So, $nR = K/T_0$.
  • Substitute $nR = K/T_0$: $Kx L_L = \left(\frac{K}{T_0}\right) \left(\frac{4T_0}{3}\right)$.
  • $Kx L_L = \frac{4K}{3} \implies x L_L = \frac{4}{3} \implies L_L = \frac{4}{3x}$.

• For the right chamber: $P_R V_R = nRT_R \implies P_R (A L_R) = nR \left(\frac{5T_0}{3}\right)$.

  • Substitute $P_R = \frac{Kx}{A}$: $\left(\frac{Kx}{A}\right) (A L_R) = nR \left(\frac{5T_0}{3}\right)$.
  • $Kx L_R = nR \left(\frac{5T_0}{3}\right)$.
  • Substitute $nR = K/T_0$: $Kx L_R = \left(\frac{K}{T_0}\right) \left(\frac{5T_0}{3}\right)$.
  • $Kx L_R = \frac{5K}{3} \implies x L_R = \frac{5}{3} \implies L_R = \frac{5}{3x}$.

4. Total Length Constraint:

• Let $L_{nat}$ be the natural length of the spring.
• Initial total length of the container: $L_{container} = L_{L,0} + (L_{nat} - x_0) + L_{R,0} = 1 + (L_{nat} - 1) + 1 = L_{nat} + 1$.
• Final total length of the container: $L_{container} = L_L + (L_{nat} - x) + L_R$.
• Since the total length of the container is constant: $L_{nat} + 1 = L_L + L_{nat} - x + L_R$ $1 = L_L - x + L_R$ $1 + x = L_L + L_R$.

5. Solve for x:

• Substitute the expressions for $L_L$ and $L_R$: $1 + x = \frac{4}{3x} + \frac{5}{3x}$ $1 + x = \frac{9}{3x}$ $1 + x = \frac{3}{x}$
• Multiply by $x$: $x(1+x) = 3$ $x + x^2 = 3$ $x^2 + x - 3 = 0$.
• Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-3)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 12}}{2}$ $x = \frac{-1 \pm \sqrt{13}}{2}$.
• Since $x$ represents compression, it must be a positive value. $x = \frac{-1 + \sqrt{13}}{2}$.

6. Find k:

• The final compression is given as $\frac{\sqrt{k-1}}{2}$.
• Therefore, $\frac{\sqrt{k-1}}{2} = \frac{\sqrt{13}-1}{2}$.
• $\sqrt{k-1} = \sqrt{13}-1$.
• Squaring both sides: $k-1 = (\sqrt{13}-1)^2$.
• $k-1 = (\sqrt{13})^2 - 2\sqrt{13}(1) + 1^2$.
• $k-1 = 13 - 2\sqrt{13} + 1$.
• $k-1 = 14 - 2\sqrt{13}$.
• $k = 15 - 2\sqrt{13}$.

This result for k does not match the form $\sqrt{k-1} = \sqrt{13}-1$ if k is expected to be an integer. Let's re-read the question carefully: "The final compression in the spring in equilibrium is given by $\frac{\sqrt{k-1}}{2}$." It's possible that the question implies $x = \frac{\sqrt{k}-1}{2}$ or $x = \frac{\sqrt{k-1}}{2}$ where k is an integer. If $x = \frac{\sqrt{k}-1}{2}$, then $\frac{\sqrt{k}-1}{2} = \frac{\sqrt{13}-1}{2}$. This would imply $\sqrt{k}-1 = \sqrt{13}-1$, which means $\sqrt{k} = \sqrt{13}$, so $k = 13$.

Let's check if the problem states $x = \frac{\sqrt{k-1}}{2}$ or $x = \frac{\sqrt{k}-1}{2}$. The image clearly shows $\frac{\sqrt{k-1}}{2}$. So $\sqrt{k-1} = \sqrt{13}-1$. This means $k-1 = (\sqrt{13}-1)^2 = 13 - 2\sqrt{13} + 1 = 14 - 2\sqrt{13}$. $k = 15 - 2\sqrt{13}$. This value of k is not an integer. Typically, such problems expect an integer value for k.

Let's assume there might be a typo in the question and it should have been $\frac{\sqrt{k}-1}{2}$. If $x = \frac{\sqrt{k}-1}{2}$, then comparing with $x = \frac{\sqrt{13}-1}{2}$, we get $\sqrt{k} = \sqrt{13}$, which means $k=13$. This is a common pattern in competitive exams where k is an integer.

Given the exact wording "$\frac{\sqrt{k-1}}{2}$", if $k$ must be an integer, there might be an issue with the question's provided form for the final compression. However, if we strictly follow the given expression: $\frac{\sqrt{k-1}}{2} = \frac{\sqrt{13}-1}{2}$ $\sqrt{k-1} = \sqrt{13}-1$ Squaring both sides: $k-1 = (\sqrt{13}-1)^2$ $k-1 = 13 - 2\sqrt{13} + 1$ $k-1 = 14 - 2\sqrt{13}$ $k = 15 - 2\sqrt{13}$

If the question intends k to be an integer, it is highly likely that the expression for compression was meant to be $\frac{\sqrt{k}-1}{2}$. In such a case, $k=13$. Without clarification, we have to assume the exact given expression. However, in the context of typical JEE/NEET questions, 'k' is usually an integer. Let's proceed with the assumption that the 'k' in the numerator is inside the square root, i.e., $\frac{\sqrt{k}-1}{2}$.

Let's re-evaluate the problem assuming the compression is $x = \frac{\sqrt{k}-1}{2}$. Then, $\frac{\sqrt{k}-1}{2} = \frac{\sqrt{13}-1}{2}$. This implies $\sqrt{k}-1 = \sqrt{13}-1$. $\sqrt{k} = \sqrt{13}$. $k = 13$.

This seems like the intended answer for an integer k.