In the virtual lab demo, a student adds particles to a container. While the pressure is held constant the initial volume of the container is 10nm. The initial temperature of the container is: . The student adds heat to the container with the particles. What part of the lab changes? When the temperature is lowered to half its temperature (150K), the volume to the initial volume. The new volume is now: . You can predict, if the temperature was increased 5x the initial temperature, then the volume would to the initial volume.

Answer 1

In this virtual lab demo, several concepts from thermodynamics are at play, particularly the ideal gas law, which relates pressure (P), volume (V), and temperature (T). The law is typically stated as:

\[ PV = nRT \]

where \( n \) is the number of moles of gas and \( R \) is the ideal gas constant.

1. Initial State: The container has a volume of \( V_1 = 10 , \text{nm}^3 \) (assuming \( \text{nm}^3 \) is the unit of volume—note that in typical scenarios, practical measurements for volumes of this scale may vary significantly) and an initial temperature (let's say it's \( T_1 \) which you will need to specify).

2. Adding Heat: When heat is added while keeping pressure constant, the volume will change according to Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature (in Kelvin).

3. Lowering Temperature: If the temperature is lowered to half of its initial value (you mention \( 150 , K \), so that would imply \( T_1 = 300 , K \)), we can use Charles's Law to find the new volume \( V_2 \):

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Substituting in the known values:

• Initial Volume, \( V_1 = 10 , \text{nm}^3 \)
• Initial Temperature, \( T_1 = 300 , K \)
• New Temperature, \( T_2 = 150 , K \)

Using the equation:

\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \]

Plugging in the numbers:

\[ V_2 = 10 , \text{nm}^3 \cdot \frac{150 , K}{300 , K} = 10 , \text{nm}^3 \cdot 0.5 = 5 , \text{nm}^3 \]

So, the new volume when the temperature is lowered to \( 150 , K \) is \( 5 , \text{nm}^3 \).

4. Increasing Temperature: If you predict that the temperature was increased to 5 times the initial temperature \( (5 \cdot T_1 = 5 \cdot 300 , K = 1500 , K) \):

\[ V_3 = V_1 \cdot \frac{T_3}{T_1} = 10 , \text{nm}^3 \cdot \frac{1500 , K}{300 , K} = 10 , \text{nm}^3 \cdot 5 = 50 , \text{nm}^3 \]

Thus, if the temperature was increased to 5 times the initial temperature, the new volume would be \( 50 , \text{nm}^3 \).

Summary:

• New volume when temperature lowered to \( 150 K \): 5 nm³
• New volume when temperature increased to \( 1500 K \): 50 nm³