Problem 14 - A Complete Resource Book in Chemistry for JEE Main

Question

By the ideal gas law the pressure of \(0.60 \mathrm{~mol} \mathrm{NH}_{3}\) gas in a \(3.00\) litre vessel at \(25^{\circ} \mathrm{C}\) is (a) \(48.9 \mathrm{~atm}\) (b) \(4.89 \mathrm{~atm}\) (c) \(0.489 \mathrm{~atm}\) (d) \(489 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The pressure is approximately 4.89 atm, so the correct answer is (b).

Step 1: Understand the Ideal Gas Law

The ideal gas law is given by the equation \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume of the gas, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. We need to apply this equation to solve for the pressure.

Step 2: Convert Temperature to Kelvin

The given temperature is \( 25^{\circ} \mathrm{C} \). To convert this to Kelvin, use the formula: \( T(K) = T(^{\circ}C) + 273.15 \). Thus, we have \( 25 + 273.15 = 298.15 \mathrm{K} \).

Step 3: Use Ideal Gas Constant

For the ideal gas constant \( R \), we can use the value \( 0.0821 \mathrm{~L \, atm \, mol^{-1} \, K^{-1}} \). This matches the units we need for pressure in atm, volume in liters, and temperature in Kelvin.

Step 4: Plug Variables into Ideal Gas Law

Insert the known values into the ideal gas law equation: \( P \times 3.00 = 0.60 \times 0.0821 \times 298.15 \).

Step 5: Solve for Pressure

Calculate the right side of the equation: \( 0.60 \times 0.0821 \times 298.15 = 14.67639 \). Now, solve for \( P \) by dividing both sides by 3.00: \[ P = \frac{14.67639}{3.00} = 4.89213 \mathrm{~atm} \].

Step 6: Determine the Closest Answer Choice

The calculated pressure is approximately \( 4.89 \mathrm{~atm} \). Comparing with the provided options, choice (b) \( 4.89 \mathrm{~atm} \) is the correct answer.

Key Concepts

Gas Pressure Calculation

Gas pressure is a measure of the force exerted by the gas particles against the walls of its container. When calculating gas pressure using the ideal gas law, we employ the formula \( PV = nRT \). Here, the pressure \( P \) is determined by rearranging the formula to isolate \( P \): \[ P = \frac{nRT}{V} \] where:

• \( n \) is the amount of gas in moles.
• \( R \) is the ideal gas constant.
• \( T \) is the temperature in Kelvin.
• \( V \) is the volume in liters.

To find gas pressure, input the appropriate values for these variables into the equation. In our exercise, we know \( n = 0.60 \text{ mol} \), \( V = 3.00 \text{ L} \), the temperature has been converted to \( 298.15 \text{ K} \), and the ideal gas constant \( R \) is \( 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \). By substituting these values into the equation, we calculate the pressure to be approximately \(4.89 \text{ atm} \). This approach is crucial for solving problems related to gases in various scientific and engineering applications.

Temperature Conversion to Kelvin

The Kelvin scale is an absolute temperature scale that is widely used in scientific calculations to ensure consistency. When dealing with thermodynamic equations like the ideal gas law, temperature must be converted to Kelvin. This is because Kelvin provides a zero reference point at absolute zero, a state where the motion of particles theoretically ceases.To convert Celsius to Kelvin, the formula used is:\[ T(K) = T(^{\circ}C) + 273.15 \]In practice, this conversion involves simply adding 273.15 to the temperature given in Celsius. For example, a temperature of \( 25^{\circ}C \) becomes \( 298.15 \text{ K} \) upon conversion. Using Kelvin in equations aligns with other SI units and helps maintain accuracy when calculating variables such as pressure or volume under changing temperatures.

Ideal Gas Constant

The ideal gas constant, denoted by \( R \), is a crucial factor in equations that relate the various properties of an ideal gas. The constant \( R \) is used in the ideal gas law formula \( PV = nRT \), effectively linking pressure, volume, temperature, and the moles of gas.Different units of \( R \) exist depending on the context:

• For pressure in \( ext{atm} \), volume in \( ext{liters} \), and temperature in \( ext{Kelvin} \), \( R \) is \( 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \).
• Other values might include \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \), often used in energy-related calculations.

In our exercise, we use \( 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \) because the problem involves calculating pressure in atmospheres and volume in liters. Selecting the correct version of \( R \) helps ensure the units cancel appropriately, leading to accurate results in calculations.