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An insulating vertical cylinder fitted with an insulating frictionless piston contains 'n' moles of an ideal diatomic gas.The initial volume and temp of the gas are Vo and To respectively,with the piston fixed tightly.the atm pressure isPo.If the piston be released,find the eq temp and vol of the gas,if the heat capacity of the cylinder and piston is negligibly small.Weight of the piston and cross sectional area of the cylinder are W and A respectively

Hrishant Goswami , 12 Years ago
Grade 10
anser 1 Answers
Askiitians Tutor Team

To solve this problem, we need to analyze the situation involving an ideal diatomic gas contained within a vertical cylinder with a piston. The key points to consider are the initial conditions of the gas, the properties of the piston, and the principles of thermodynamics that govern the behavior of gases. Let's break this down step by step.

Initial Conditions and Setup

We start with 'n' moles of an ideal diatomic gas at an initial volume \( V_0 \) and temperature \( T_0 \). The piston is tightly fixed, and the atmospheric pressure is \( P_0 \). When the piston is released, the gas will expand against the atmospheric pressure until it reaches equilibrium.

Understanding the Process

Since the cylinder and piston are insulating, we can assume that no heat is exchanged with the surroundings. This means that the process will be adiabatic, and we will use the principles of adiabatic expansion to find the final temperature and volume of the gas.

Applying the Ideal Gas Law

The ideal gas law states that:

  • PV = nRT

Where:

  • P = pressure
  • V = volume
  • n = number of moles
  • R = universal gas constant
  • T = temperature

Equilibrium Conditions

When the piston is released, the gas will expand until the pressure inside the cylinder equals the atmospheric pressure \( P_0 \). At equilibrium, we have:

  • Final Pressure, \( P = P_0 \)

Final Volume and Temperature

Using the ideal gas law at equilibrium, we can express the final volume \( V_f \) as:

  • \( P_0 V_f = nRT_f \)

Rearranging gives us:

  • \( V_f = \frac{nRT_f}{P_0} \)

Adiabatic Process for Diatomic Gas

For an adiabatic process involving an ideal diatomic gas, we can use the relation:

  • \( TV^{\gamma - 1} = \text{constant} \)

Where \( \gamma \) (gamma) is the heat capacity ratio, which for a diatomic gas is approximately \( \frac{7}{5} \) or 1.4. Thus, we can write:

  • \( T_0 V_0^{\gamma - 1} = T_f V_f^{\gamma - 1} \)

Solving for Final Temperature

Substituting \( V_f \) into the adiabatic equation gives:

  • \( T_0 V_0^{\gamma - 1} = T_f \left( \frac{nRT_f}{P_0} \right)^{\gamma - 1} \)

This equation can be rearranged to solve for \( T_f \). However, since we have two unknowns, we can also express \( T_f \) in terms of \( V_f \) and \( P_0 \) using the ideal gas law.

Final Steps

To find \( T_f \) explicitly, we can combine these equations. After some algebra, we can derive:

  • \( T_f = \frac{T_0 V_0^{\gamma - 1} P_0^{1 - \gamma}}{nR} \)

And substituting back to find \( V_f \) will yield the final volume after the piston has moved to equilibrium.

Conclusion

In summary, by applying the ideal gas law and the principles of adiabatic processes, we can determine the final temperature and volume of the gas after the piston is released. The key takeaway is that the gas expands until its pressure equals the atmospheric pressure, and this expansion is governed by the relationships defined by the ideal gas law and the adiabatic process equations.

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