Problem 52 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The pressure exerted by \(12 \mathrm{~g}\) of an ideal gas at temperature \(t^{\circ} \mathrm{C}\) in a vessel of volume \(V\) litre is one atm. When the temperature is increased by 10 degrees at the same volume, the pressure increases by \(10 \%\). Calculate the temperature \(t\) and volume \(V\). (Molecular weight of the gas \(=120\).)

Step-by-Step Solution

Answer

The temperature \( t = -173.15 \, ^{\circ}C \) and volume \( V = 0.821 \, \text{litres} \).

Step 1: Examine the initial conditions

The initial pressure (P) is one atm, the mass of the gas (m) is 12 g, and the molecular weight (M) is 120 g/mol. We can use the ideal gas law in the form \( PV = nRT \) to find n (the number of moles), where \(n = \frac{m}{M}\). Substituting the values: \( n = \frac{12}{120} = 0.1 \) mols.

Step 2: Apply the initial ideal gas equation

Using \( P_1V = nRT_1 \) where \( P_1 = 1 \) atm and \( T_1 = t + 273.15 \) K, we have \( 1 \cdot V = 0.1 \cdot R \cdot (t + 273.15) \). Thus, the equation becomes \( V = 0.1R(t + 273.15) \). This equation relates the volume \( V \) to the initial temperature \( t \).

Step 3: Analyze the condition after temperature increase

When the temperature increases by 10 degrees, the new temperature is \( T_2 = (t + 10) + 273.15 \) K. The new pressure \( P_2 \) is 10% more than the initial pressure, so \( P_2 = 1.1 \) atm. Apply the ideal gas equation: \( 1.1 \cdot V = 0.1 \cdot R \cdot (t + 283.15) \).

Step 4: Set up and solve the equations

We have two equations: 1. \( V = 0.1R(t + 273.15) \)2. \( 1.1V = 0.1R(t + 283.15) \).Substitute the expression for \( V \) from the first equation into the second equation:\[ 1.1(0.1R(t + 273.15)) = 0.1R(t + 283.15) \].Simplify this equation by canceling \( 0.1R \) on both sides:\[ 1.1(t + 273.15) = t + 283.15 \].

Step 5: Solve for temperature \( t \)

Expand and simplify the equation:\[ 1.1t + 300.465 = t + 283.15 \].Rearrange terms to isolate \( t \):\[ 0.1t = 283.15 - 300.465 \]\[ 0.1t = -17.315 \].Solve for \( t \):\[ t = -17.315 / 0.1 = -173.15 \,^{\circ}C \].

Step 6: Find the volume \( V \)

Using the expression for \( V \) from Step 2 and the value of \( t = -173.15 \):\[ V = 0.1R(-173.15 + 273.15) \].Substitute \( R = 0.0821 \) L atm / K mol for the ideal gas constant:\[ V = 0.1 \times 0.0821 \times 100 \].\[ V = 0.821 \text{ litres} \].

Key Concepts

Moles Calculation

Calculating the number of moles is a foundational concept in chemistry and pivotal when using the Ideal Gas Law. To find the moles (), you need the mass of the gas and its molecular weight. The formula is given by:

• \[ n = \frac{m}{M} \]

In this formula:

• \( m \) is the mass of the gas in grams.
• \( M \) is the molecular weight of the gas in grams per mole.

The well-known process involves dividing the mass of the gas (12 g in this case) by its molecular weight (120 g/mol).
This gives:

• \[ n = \frac{12}{120} = 0.1 \] moles.

This calculation of moles helps in further determining the relationships in the equation of state, such as calculating pressure and volume, through the Ideal Gas Law formula.

Temperature Conversion

In chemistry problems involving gases, temperatures must be in the Kelvin scale. This is because the Ideal Gas Law requires absolute temperature for accurate results. To convert a Celsius temperature to Kelvin, you simply add 273.15.

• \[ T = t + 273.15 \]

Here, \( T \) is the temperature in Kelvin and \( t \) is the temperature in Celsius.
In our example, if the initial temperature is unknown and denoted as \( t \) degrees Celsius, the Kelvin conversion gives \( T_1 = t + 273.15 \). When the temperature is increased by 10 degrees, the new temperature becomes \( T_2 = (t + 10) + 273.15 \) K.
Kelvin is preferred as it ensures no zero or negative values, which can cause undefined or nonsensical results in calculations involving volume or pressure. This conversion underlies crucial aspects of consistency and integrity in thermodynamic equations.

Volume Calculation

Volume calculation in gas laws relies heavily on relationships established by the Ideal Gas Law: \( PV = nRT \).
The challenge is to find the volume \( V \) using known quantities of pressure \( P \), moles \( n \), and the gas constant \( R \). From the Ideal Gas Law, rearranging it to solve for \( V \) gives:

• \[ V = \frac{nRT}{P} \]

For our problem, with the pressure set to 1 atm and an established relationship involving the temperature \( T_1 \), the initial expression becomes:

• \[ V = 0.1R(t + 273.15) \]

This captures the volume in terms of unknown temperature. After increasing temperature and pressure, you set up a second equation:

• \[ 1.1V = 0.1R(t + 283.15) \]

Using these expressions helps solve for both temperature and volume by eliminating variables and logically progressing through calculations. The final solution involves substituting known values into these expressions to determine the precise volume in liters, namely \( V \approx 0.821 \) L, once simpler calculations isolate the temperature \( t \).