Question

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure changes from 760 torr to \(202 \mathrm{kPa}\), and the temperature changes from \(37^{\circ} \mathrm{C}\) to \(155 \mathrm{~K}\) (b) The temperature changes from \(305 \mathrm{~K}\) to \(32^{\circ} \mathrm{C},\) and the pressure changes from 2 atm to \(101 \mathrm{kPa}\). (c) The pressure is reduced by a factor of 4 (at constant \(T\) ).

Short answer

a) Volume decreases to 6.38 L from 25.47 L.b) Volume increases to 24.98 L from 12.52 L.c) Volume increases by a factor of 4.

Step-by-step solution

- Understand the Ideal Gas Law

The Ideal Gas Law is given by the equation \[PV = nRT\], where: - \(P\) is the pressure in atmospheres, - \(V\) is the volume in liters, - \(n\) is the number of moles, - \(R\) is the universal gas constant (0.0821 atm·L/mol·K), - \(T\) is the temperature in Kelvin.

- Convert Given Values to Consistent Units (Part a)

Convert the given pressures and temperatures to consistent units (atm and K): \[202 \text{ kPa} = 202 \times 0.00986923 \text{ atm} = 1.993 \text{ atm} \] \[37^{\text{ o}} \text{C} = 37 + 273.15 = 310.15 \text{ K} \] \[155 \text{ K} (No conversion needed) \]

- Apply Ideal Gas Law (Part a)

Using the Ideal Gas Law equation for both initial and final states: Initial: \(P_1V_1 = nRT_1 \rightarrow 1 \text{ atm} \times V_1 = 1 \text{ mol} \times 0.0821 \text{ atm·L/mol·K} \times 310.15 \text{ K}\) \[ V_1 = 25.47 \text{ L} \] Final: \(P_2V_2 = nRT_2 \rightarrow 1.993 \text{ atm} \times V_2 = 1 \text{ mol} \times 0.0821 \text{ atm·L/mol·K} \times 155 \text{ K}\) \[ V_2 = 6.38 \text{ L} \]

- Convert Given Values to Consistent Units (Part b)

Convert the given pressures and temperatures to consistent units (atm and K): \[32^{\text{ o}} \text{C} = 32 + 273.15 = 305.15 \text{ K}\] \[101 \text{ kPa} = 101 \times 0.00986923 \text{ atm} = 0.998 \text{ atm}\] The initial temperature and pressure are already in consistent units: 305 K and 2 atm.

- Apply Ideal Gas Law (Part b)

Using the Ideal Gas Law equation for both initial and final states: Initial: \(P_1V_1 = nRT_1 \rightarrow 2 \text{ atm} \times V_1 = 1 \text{ mol} \times 0.0821 \text{ atm·L/mol·K} \times 305 \text{ K}\) \[ V_1 = 12.52 \text{ L} \] Final: \(P_2V_2 = nRT_2 \rightarrow 0.998 \text{ atm} \times V_2 = 1 \text{ mol} \times 0.0821 \text{ atm·L/mol·K} \times 305.15 \text{ K}\) \[ V_2 = 24.98 \text{ L} \]

- Apply the Pressure Reduction Factor (Part c)

For this scenario, the pressure is reduced by a factor of 4 while the temperature is held constant: Initially: \(PV = nRT\) means that \(P_1V = nRT \). When the pressure is reduced by a factor of 4: \(\frac{P_1}{4} \). According to the Ideal Gas Law, keeping \(T\) constant: \(\frac{P_1}{4} \times V_2 = nRT \). Since \(nRT = P_1V\): \[P_1V = \frac{P_1}{4} \times V_2 \rightarrow 1V = \frac{V_2}{4} \times 4\rightarrow V_2 = 4V_1 \]. Thus, the volume would increase by a factor of 4.

Key concepts

Pressure Conversion

In the context of the Ideal Gas Law, pressure is often measured in different units such as atmospheres (atm), kilopascals (kPa), or torr. To successfully apply the Ideal Gas Law, all pressures must be in the same units.

For example, converting from kPa to atm requires multiplying by 0.00986923, as seen in the conversion: 202 kPa = 202 × 0.00986923 atm = 1.993 atm.

Converting from units like torr to atm involves dividing by 760, since 1 atm equals 760 torr. Ensuring consistent pressure units is essential for accurate calculations using the Ideal Gas Law.

Temperature Conversion

Temperature must be in Kelvin (K) when applying gas laws, as the Kelvin scale starts at absolute zero.

Converting from Celsius to Kelvin is straightforward: add 273.15 to the Celsius temperature. For instance, converting 37°C to Kelvin results in: 37 + 273.15 = 310.15 K.

Conversely, converting back to Celsius involves subtracting 273.15 from the Kelvin temperature. Proper temperature conversion is vital for the Ideal Gas Law to work accurately, ensuring the correct relationship between pressure, volume, and temperature.

Gas Constant

The universal gas constant, denoted as R, plays a critical role in the Ideal Gas Law. Its value is 0.0821 atm·L/mol·K. This constant allows the bridging of units like atmospheres, liters, moles, and Kelvin in the equation PV = nRT.

R ensures that the relationship between pressure, volume, and temperature holds true for one mole of an ideal gas. Regardless of the state of the gas concerning pressure and temperature, applying R consistently ensures that calculations yield correct and meaningful results.

Effect on Volume

Understanding the effect of changes in pressure and temperature on the volume of a gas is crucial when using the Ideal Gas Law. According to the equation PV = nRT, volume varies directly with temperature and inversely with pressure.

For instance, if pressure is reduced by a factor of 4 (keeping temperature constant), the volume increases by a factor of 4. This is shown mathematically as:

Initially: PV = nRT

Reduced Pressure: P/4 * V2 = nRT. Since nRT = P * V1, it follows that V2 = 4 * V1.

Similarly, changes in temperature like from 310.15 K to 155 K, with other factors constant, would lead to a decrease in volume, emphasizing the direct relationship between temperature and volume.