Problem 80 - Chemistry: Principles and Reactions

Question

Tank A has ammonis at \(300 \mathrm{~K}\), Tank B hes nitrogen gas at \(150 \mathrm{~K}\). Thaks A and B have the same volume. Compare the presrures in tanks \(A\) and \(B\) if (a) tank B has twice as many moles of nitrogen as tank \(A\) has of ammonia. (b) tank A has the same number of moles of ammonia as tank \(\mathrm{B}\) has of nitrogen. (Try to do this without a calculatorl)

Step-by-Step Solution

Answer

Question: Compare the pressures in tank A and tank B under the two given conditions. Answer: For condition (a), the pressure in tank A is half that of tank B. For condition (b), the pressure in tank A is twice that of tank B.

Step 1: Tank A pressure formula

Using the Ideal Gas Law, \(PV = nRT\), we first note that the Volume (V) and the ideal gas constant (R) are the same for both tanks. So, we can write the pressure and temperature relationship for tank A as: \(P_A = \frac{n_{NH_3}R}{V} * T_{NH_3}\) where \(P_A\) is the pressure in tank A, \(n_{NH_3}\) is the number of moles of ammonia, and \(T_{NH_3} = 300 \mathrm{K}\).

Step 2: Tank B pressure formula

Similarly, for tank B, we can write the relationship between pressure, temperature, and moles of nitrogen as: \(P_B = \frac{n_{N_2}R}{V} * T_{N_2}\) where \(P_B\) is the pressure in tank B, \(n_{N_2}\) is the number of moles of nitrogen, and \(T_{N_2} = 150 \mathrm{K}\). Now let's find the ratios of the pressures in tanks A and B for conditions (a) and (b).

Step 3: Pressure ratio for condition (a)

For condition (a), we are given that the number of moles of nitrogen in tank B is twice that of ammonia in tank A: \(n_{N_2} = 2n_{NH_3}\). Now, let us find the ratio of \(P_A\) to \(P_B\): \(\frac{P_A}{P_B} = \frac{(\frac{n_{NH_3}R}{V} * 300)}{(\frac{(2n_{NH_3})R}{V} * 150)} = \frac{300}{2 * 150} = \frac{1}{2}\) So, in condition (a), the pressure in tank A is half that of tank B.

Step 4: Pressure ratio for condition (b)

Condition (b) states that the number of moles of ammonia in tank A is equal to the number of moles of nitrogen in tank B: \(n_{NH_3} = n_{N_2}\). The ratio of pressures of A and B is: \(\frac{P_A}{P_B} = \frac{(\frac{n_{NH_3}R}{V} * 300)}{(\frac{n_{NH_3}R}{V} * 150)} = \frac{300}{150} = 2\) Hence, for condition (b), the pressure in tank A is twice that of tank B.

Key Concepts

Pressure-Temperature Relationship

The relationship between pressure and temperature of a gas is central to understanding the behavior of gases, especially in contexts like that of an enclosed tank. This is often explored through the Ideal Gas Law, given by the formula: \[ PV = nRT \]where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant,
• \( T \) is the temperature in Kelvin.

In a situation where the volume and the number of moles are constant, the pressure is directly proportional to the temperature. This means that as temperature increases, pressure also increases if other conditions remain the same. Conversatively, with cooler temperatures, pressure tends to drop.
In the given exercise, comparing two tanks, we can see how the pressure differs in Tanks A and B due to their differing temperatures. Keeping the same conditions of volume and ideal gas constant, it’s evident that Tank A at \(300 \mathrm{~K}\) will have a different pressure reading compared to Tank B at \(150 \mathrm{~K}\). This understanding is vital when considering the state changes of gases in various temperatures.

Moles and Gas Calculations

To dive deeper into gas calculations, it is crucial to understand the concept of moles. A mole is a standard scientific unit for measuring large quantities of very small entities such as atoms or molecules. In the context of the Ideal Gas Law, the number of moles of gas directly affects the relationship between pressure, volume, and temperature.
The exercise explores scenarios where the number of moles changes the resulting pressure in tanks. For condition (a), Tank B has double the moles of nitrogen compared to ammonia in Tank A. The relationship for calculating pressure simplifies when we use ratios, helping predict outcomes without extensive calculations. With twice the moles at half the temperature, Tank B's pressure compensates, showing higher pressure due to more moles.
Condition (b) simplifies further as both tanks hold the same number of moles. This creates a direct comparison of pressure based purely on their temperature relationship, thereby making it straightforward to resolve pressure differences as they relate to temperature variances.

Comparative Analysis of Gas Conditions

A comparative analysis of gas conditions involves looking at all factors like molecule number, temperature, and volume to understand their combined impact on gas behavior. The exercise helps to assess these conditions succinctly by posing scenarios for calculation without needing a calculator.

• In condition (a), with double the moles but half the temperature in Tank B, it leads to a greater pressure compared to Tank A. This affirms that the effect of increased moles surpasses the reduced impact of lower temperature.
• For condition (b), with equal moles in both tanks, temperature becomes the decisive factor. Tank A, being at a higher temperature, exhibits a greater pressure than Tank B, showing how a minor change in temperature can dramatically affect pressure given constant volume and moles.

This comparative analysis is essential for predicting gas behavior under varying conditions, such as in industrial applications or understanding natural processes in atmospheric sciences. Understanding these concepts fully can make solving gas-related problems straightforward and intuitive.