Problem 92 - Master Resource Book in JEE Main Physics

Question

An inflated rubber balloon contains one mole of an ideal gas, has a pressure \(p\), volume \(V\) and temperature \(T\). If the temperature rises to \(1.1 T\), and the volume is increased to \(1.05 \mathrm{~V}\). the final pressure will be (a) \(1.1 p\) (b) \(p\) (c) less than \(p\) (d) between \(p\) and \(1.1 p\)

Step-by-Step Solution

Answer

The final pressure is between \( p \) and \( 1.1p \).

Step 1: Identify the Initial Conditions

We begin with a balloon containing one mole of an ideal gas at initial conditions of pressure \( p \), volume \( V \), and temperature \( T \). Given these are initial states, we can use the ideal gas law: \( pV = nRT \) where \( n = 1 \).

Step 2: Understand Changes in Conditions

The problem states that the temperature is increased to \( 1.1 T \) and the volume is increased to \( 1.05 V \). We need to determine how these changes affect the pressure.

Step 3: Apply the Ideal Gas Law to the Final Conditions

Using the ideal gas law, the final conditions can be expressed as \( p' \times 1.05V = 1 \times R \times 1.1T \). This simplifies to \( p' = \frac{1.1RT}{1.05V} \).

Step 4: Relate New Pressure to Initial Pressure

We know that the initial condition was \( p = \frac{RT}{V} \). Thus, the final pressure \( p' \) can be related back to the initial pressure: \( p' = \frac{1.1}{1.05} \times p \). Solving this, we get \( p' \approx 1.0476p \).

Step 5: Determine the Range of the Final Pressure

Comparing \( p' \approx 1.0476p \) with the given choices: \( p' \) is greater than \( p \) but less than \( 1.1p \). Hence, the final pressure \( p' \) is between \( p \) and \( 1.1p \).

Key Concepts

Understanding Pressure in Ideal Gases

Pressure is a measure of the force that gas molecules exert when they collide with the walls of their container. In the context of an ideal gas, pressure is related to the number of gas molecules, their speed, and how frequently they hit the container's walls.
For the ideal gas law, pressure (\( p \)) is mathematically represented in the equation \( pV = nRT \), where \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
The pressure can change if any of these variables change. In the exercise scenario, when both volume and temperature change, so does the pressure. To find how much the pressure changes, we rearrange the ideal gas equation based on the new conditions provided in the problem, considering the constant number of moles. Hence, understanding pressure is key to solving problems involving ideal gases.

Exploring Volume of Ideal Gases

Volume is the space that the gas occupies. In our balloon example, volume is crucial as it directly affects pressure alongside temperature.\( V \), or volume, is part of the equation \( pV = nRT \), and it's important to see how changes in volume influence other properties like pressure.
When the volume of a gas increases (as it did in this exercise from \( V \) to \( 1.05V \)), it means there's more space for the gas molecules to move around. This increase typically results in a reduction in pressure, assuming temperature is constant (Boyle's Law).
However, in this exercise, both volume and temperature increased. This requires a holistic understanding of how these factors work together in equations such as \( p' \times 1.05V = 1 \times R \times 1.1T \). The key takeaway here is understanding how volume changes impact the pressure and the interplay with other variables in the ideal gas law.

The Role of Temperature in Ideal Gases

Temperature is a measure of the average kinetic energy of gas particles. In ideal gas calculations, temperature must always be measured in Kelvin for accuracy. It plays a direct role in the ideal gas equation \( pV = nRT \).
Increased temperature means more kinetic energy for the gas molecules, causing them to move faster. This, in turn, can increase pressure if the volume is held constant. However, in this exercise, both temperature and volume change.
The temperature had risen to \( 1.1T \), and this change impacts pressure when related through the ideal gas law. The equation simplifies these relations: the increase in temperature to \( 1.1T \) contributes to increasing the pressure if volume changes are considered. Hence, when calculating final pressure, the temperature's role is significant and must be balanced against volume changes to understand its full impact on pressure.