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 of the gas is 4.89 atm, option (b).

Step 1: Write down the Ideal Gas Law Formula

The ideal gas law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

Since the temperature given is in degrees Celsius, convert it to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \). For this problem, \( T = 25 + 273.15 = 298.15 \) K.

Step 3: Substitute Known Values

Substitute the known values into the ideal gas law. Here, \( n = 0.60 \) mol, \( V = 3.00 \) L, and \( T = 298.15 \) K. The ideal gas constant \( R \) is \( 0.0821 \) L·atm/mol·K.

Step 4: Solve for Pressure (P)

Rearrange the ideal gas law to solve for pressure: \( P = \frac{nRT}{V} \). Substitute the values: \( P = \frac{(0.60)(0.0821)(298.15)}{3.00} \). Calculate to find \( P \approx 4.89 \) atm.

Step 5: Compare Calculated Pressure with Choices

The calculated pressure is \( 4.89 \) atm. Compare this with the given options: (a) 48.9 atm, (b) 4.89 atm, (c) 0.489 atm, (d) 489 atm. Option (b) is the correct answer.

Key Concepts

Gas Pressure Calculation

When dealing with gases, calculating pressure is a key aspect, and it becomes simple with the Ideal Gas Law. The law is expressed as \( PV = nRT \), where \( P \) means pressure, \( V \) is volume, \( n \) refers to moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. The equation allows us to determine one parameter if the others are known.

In our exercise, we need to find the pressure of the gas: Rearrange the equation to \( P = \frac{nRT}{V} \). This step is crucial because it directly links the pressure of a gas to its amount, volume, and temperature under a constant. Here, the formula makes it easy to substitute the known values for a quick and accurate calculation. Remember that each component of this equation serves a distinct purpose, making it fundamental in chemistry.

Temperature Conversion to Kelvin

To use the Ideal Gas Law, temperatures must be in Kelvin because the Kelvin scale is absolute. Converting Celsius to Kelvin is straightforward: just add 273.15 to the temperature in Celsius.

For example, if our temperature is 25°C, the conversion would go as follows: \( T(K) = 25 + 273.15 = 298.15 \) K. This step is important because calculations with the Ideal Gas Law are only accurate with temperature in Kelvin, as it sets zero as absolute zero.

Always ensure the temperature is converted before substituting into any gas law equation. This simple addition transforms Celsius, our everyday temperature measure, into a form used worldwide in scientific study.

Ideal Gas Constant Value

The Ideal Gas Constant, \( R \), is an integral part of the Ideal Gas Law. It acts as a proportional constant and its value is 0.0821 L·atm/mol·K. This value is chosen because it aligns with the units we commonly use in gas law calculations: liters for volume, atmospheres for pressure, moles for the amount of substance, and Kelvin for temperature.

Understanding \( R \) is crucial for solving problems accurately. This constant ensures that when we plug our values into the equation \( PV = nRT \), everything balances perfectly in terms of units.

Different forms of \( R \) exist depending on the units of pressure, such as mmHg, but for our problem, it's vital to use the one matching the units provided. Keeping the right value and units in mind prevents mistakes and leads to correct calculations.