To solve this problem, we need to analyze the relationships given for the gas's volume $V$, temperature $T$, and pressure $P$.

Problem Breakdown and Solution

The problem states that:

• $V$ varies directly with $T$, meaning $V \propto T$.
• $V$ varies inversely with $P$, meaning $V \propto \frac{1}{P}$.

This relationship can be written as: $V = k \frac{T}{P}$ where $k$ is a constant.

Step 1: Find the Constant $k$

We are given that:

• $V = 42 \, \text{cm}^3$
• $T = 84 \, \text{K}$
• $P = 8 \, \frac{\text{kg}}{\text{cm}^2}$

Substitute these values into the formula to solve for $k$: $42 = k \frac{84}{8}$ Simplify: $42 = k \times 10.5$ $k = \frac{42}{10.5} = 4$

Step 2: Find the Pressure $P$ When $T = 185 \, \text{K}$ and $V = 74 \, \text{cm}^3$

Now that we know $k = 4$, we can use the equation: $V = 4 \frac{T}{P}$ Substitute $V = 74 \, \text{cm}^3$ and $T = 185 \, \text{K}$: $74 = 4 \frac{185}{P}$ Rearrange to solve for $P$: $P = 4 \frac{185}{74}$ Calculate the value of $P$: $P = 10 \, \frac{\text{kg}}{\text{cm}^2}$

Answer

The pressure $P$ when $T = 185 \, \text{K}$ and $V = 74 \, \text{cm}^3$ is: $P = 10 \, \frac{\text{kg}}{\text{cm}^2}$

Would you like a deeper explanation of any part? Here are some related questions that could expand your understanding:

1. How does the direct and inverse relationship affect the volume in other gas law problems?
2. What are other real-world applications of this type of relationship between pressure, temperature, and volume?
3. How would the answer change if $V$ and $T$ were in different units, such as liters and Celsius?
4. Can this relationship be derived from the Ideal Gas Law?
5. How would the equation change if we considered a variable amount of gas (non-fixed)?

Tip: When solving problems involving proportional relationships, setting up the equation properly according to direct and inverse variations is key to finding the correct constant and solution.