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Chapter 8: Problem 1108

A monoatomic ideal gas, intially at temperature \(1_{1}\) is enclosed in a cylinders fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature \(\mathrm{T}_{2}\) by releasing the piston suddenly If \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\) the lengths of the gas column before and after expansion respectively, then then \(\left(\mathrm{T}_{1} / \mathrm{T}_{2}\right)\) is given by (A) \(\left\\{\mathrm{L}_{1} / \mathrm{L}_{2}\right\\}^{(2 / 3)}\) (B) \(\left\\{\mathrm{L}_{2} / \mathrm{L}_{1}\right\\}^{(2 / 3)}\) (C) \(\left\\{\mathrm{L}_{1} / \mathrm{L}_{2}\right\\}\) (D) \(\left\\{\mathrm{L}_{2} / \mathrm{L}_{1}\right\\}\)

Short Answer

Expert verified
The short answer to the question is: (A) \(\left\\{\frac{L_1}{L_2}\right\\}^{(2/3)}\).

Step by step solution

01

Understand the adiabatic process

During an adiabatic process, the gas expands without exchanging heat with its surroundings. In such a process, any change in the internal energy of the gas comes solely from the work done by or on the gas. For a monoatomic ideal gas, the adiabatic process can be described by the equation: \(PV^{\gamma} = k\), where P is the pressure, V is the volume, γ (gamma) is the adiabatic index, and k is a constant. The adiabatic index, γ, for a monoatomic ideal gas is (5/3).
02

Replace volume with length variables

We are given the relationship between the lengths of the gas column before and after expansion, L1 and L2. We can assume the cross-sectional area of the cylinder remains constant. The volume before and after expansion can be expressed as Area × Length. Since the cross-sectional area of the cylinder is the same during expansions, we can simplify the relationship as: \(V_1 = A \times L_1\) \(V_2 = A \times L_2\) We can then substitute these expressions for V1 and V2 in the adiabatic process equation for initial and final states: \(P_1 (A \times L_1)^{5/3} = k\) \(P_2 (A \times L_2)^{5/3} = k\)
03

Eliminate pressure terms

We can eliminate the pressure P1 and P2 by using the ideal gas law: \(P_1 V_1 = n R T_1\) \(P_2 V_2 = n R T_2\) By dividing the equations, we can get a relation between the two pressures: \(\frac{P_1}{P_2} = \frac{T_1}{T_2}\) Now, divide the adiabatic process equation for the initial state by the adiabatic process equation for the final state: \(\frac{P_1 (A \times L_1)^{5/3}}{P_2 (A \times L_2)^{5/3}} = 1\) Substitute the pressure ratio: \(\frac{T_1}{T_2} = \left(\frac{L_1}{L_2}\right)^{5/3}\)
04

Find the required ratio

Now we need to find the ratio \(\left(\frac{T_1}{T_2}\right)\), which can be found by finding \(\left(\frac{L_1}{L_2}\right)^{2/3}\) from the equation we derived in Step 3. \(\frac{T_1}{T_2} = \left(\frac{L_1}{L_2}\right)^{5/3}\) Raise both sides to the power of (3/5) to eliminate the (5/3) power: \(\left(\frac{T_1}{T_2}\right)^{3/5} = \left(\frac{L_1}{L_2}\right)\) Now, raise both sides to the power of (2/3) to find the ratio we need: \(\left(\frac{T_1}{T_2}\right) = \left(\frac{L_1}{L_2}\right)^{2/3}\) This matches the option (A), so the correct answer is: (A) \(\left\\{\frac{L_1}{L_2}\right\\}^{(2/3)}\)

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Most popular questions from this chapter

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature $55^{\circ} \mathrm{C}$. Assume no heat is lost to the surroundings and the pressure in the container is constant atm. What is the final temperature the System? (A) \(72^{\circ} \mathrm{C}\) (B) \(48^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(94^{\circ} \mathrm{C}\)

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

What is an adiabatic Bulk modulus of hydrogen gas at NTP? (A) \(1.4\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (B) \(1.4 \times 10^{5}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (C) \(1 \times 10^{-8}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (D) \(1 \times 10^{5}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\)

One mole of \(\mathrm{O}_{2}\) gas having a Volume equal to \(22.4\) liter at \(\mathrm{O}\) \(c\) and 1 atmospheric Pressure is Compressed isothermally so that its volume reduces to \(11.2\) liters. The work done in this Process is (A) \(1672.4 \mathrm{~J}\) (B) \(-1728 \mathrm{~J}\) (C) \(1728 \mathrm{~J}\) (D) \(-1572.4 \mathrm{~J}\)
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