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Chapter 5: Problem 39

If pressure of 2 mol of an ideal gas at \(546 \mathrm{~K}\) having volume \(44.8\) litre is: (a) \(2 \mathrm{~atm}\) (b) \(3 \mathrm{~atm}\) (c) \(4 \mathrm{~atm}\) (d) \(\mathrm{I} \mathrm{atm}\)

Short Answer

Expert verified
The pressure of the ideal gas is 2 atm.

Step by step solution

01

Identify the given values

We are given the amount of gas, number of moles \( n = 2 \) mol, temperature \( T = 546 \ \text{K} \), and volume \( V = 44.8 \ \text{liters} \). We need to determine the pressure of this ideal gas using the ideal gas equation.
02

Recall the Ideal Gas Law equation

The Ideal Gas Law is given by the equation \( PV = nRT \), 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.
03

Choose the ideal gas constant

For this problem, we will use the ideal gas constant \( R = 0.0821 \ \text{L atm mol}^{-1} \text{K}^{-1} \) suitable for pressure in atm, volume in liters, and temperature in Kelvin.
04

Substitute given values into the equation

Substitute \( n = 2 \) mol, \( V = 44.8 \ \text{liters} \), \( R = 0.0821 \ \text{L atm mol}^{-1} \text{K}^{-1} \), and \( T = 546 \ \text{K} \) into the formula. We need to solve for \( P \), thus: \[ P = \frac{nRT}{V} = \frac{2 \times 0.0821 \times 546}{44.8} \]
05

Perform the calculation

Calculate the right side of the equation: 1. Multiply the values: \( 2 \times 0.0821 \times 546 = 89.6682 \).2. Divide the product by the volume: \( \frac{89.6682}{44.8} = 2 \ \text{atm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Let's dive into how we calculate the pressure of an ideal gas using the Ideal Gas Law. Pressure is a measure of force exerted by the gas molecules as they collide with the walls of their container. The Ideal Gas Law provides a straightforward way to compute this pressure. The formula is given by:
  • \( PV = nRT \)
Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
To find the pressure \( P \), you rearrange the Ideal Gas Law formula to:
  • \( P = \frac{nRT}{V} \)
Substitute the known values into this equation and compute. Remember, getting pressure right requires using consistent units that match the gas constant you choose.
Temperature in Kelvin
In the Ideal Gas Law equation, it's crucial to use temperature in Kelvin. The Kelvin scale is an absolute temperature scale, meaning it starts from absolute zero, the point where all molecular motion stops. This property makes it perfect for gas law calculations.
To convert a temperature from Celsius to Kelvin, use the formula:
  • \( T( ext{K}) = T(掳 ext{C}) + 273.15 \)
Using Kelvin in equations ensures that the direct proportionality between temperature and pressure (and volume) holds true as described by Charles's Law.
For example, if you have a problem where the temperature given is in degrees Celsius, don鈥檛 forget to convert it to Kelvin before plugging it into the Ideal Gas Law.
Gas Constant
The gas constant \( R \) in the Ideal Gas Law is a crucial element, acting as a bridge between the macroscopic properties of gases. Its value depends on the units used for pressure, volume, and temperature. For our calculation with pressure in atmospheres (atm), volume in liters, and temperature in Kelvin, \( R \) is:
  • 0.0821 \( ext{L atm mol}^{-1} ext{K}^{-1} \)
Choosing the appropriate \( R \) value is essential, as it ensures all the units in the equation are compatible and will allow you to correctly calculate the desired variable. For different problems, \( R \) can also be expressed in other units, such as joules per mole Kelvin (\( ext{J mol}^{-1} ext{K}^{-1} \)), depending on whether the pressure is given in pascals or other units. This flexibility makes \( R \) a universal constant in chemistry, adaptable to your specific needs during calculations.

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Most popular questions from this chapter

The density of a gas is \(1.964 \mathrm{~g} \mathrm{dm}^{-3}\) at \(273 \mathrm{~K}\) and \(76 \mathrm{~cm} \mathrm{Hg}\). The gas is: (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{CO}_{2}\) (d) \(\mathrm{Xe}\)

An ideal gas obeying kinetic gas equation can be liquefied if: (a) It cannot be liquefied at any value of \(\mathrm{P}\) and \(\mathrm{T}\) (b) Its temperature is more than Boyle's temperature (c) Its temperature is more than critical temperature (d) Its pressure is more than critical pressure

\(0.24 \mathrm{~g}\) of a volatile gas upon vaporization gives \(45 \mathrm{~mL}\) vapour at NTP. What will be the vapour density of the substances? (density of \(\mathrm{H}_{2}=1\) ): (a) \(95.39\) (b) \(5.973\) (c) \(95.93\) (d) \(59.73\)

The root mean square velocity of one mole of a monoatomic gas having molar mass \(\mathrm{M}\) is \(\mathrm{u}_{\mathrm{rms}}\) - The relation between the average kinetic energy (E) of the gas and \(\mathrm{u}_{\mathrm{rms}}\) is: (a) \(\mathrm{u}_{\mathrm{rms}}=\sqrt{(3 \mathrm{E} / 2 \mathrm{M})}\) (b) \(\mathrm{u}_{\mathrm{mas}}=\sqrt{(2 \mathrm{E} / 3 \mathrm{M})}\) (c) \(\mathrm{u}_{\mathrm{rms}}=\sqrt{(2 \mathrm{E} / \mathrm{M})}\) (d) \(\mathrm{u}_{\mathrm{mas}}=\sqrt{(\mathrm{E} / 3 \mathrm{M})}\)

Which one of the following statement is not true about the effect of an increase in temperature on the distribution molecular speeds in a gas? (a) The most probable speed increases (b) The fraction of the molecules with the most probable speed increases (c) The distribution becomes broader (d) The area under the distribution curve remains the same as the under the lower temperature
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