A 3.50L sample of neon gas has a pressure of 0.950 atm at 20 degrees C. What would the volume be if the pressure increased to 1.50 atm and the temperature remained constant?

I have the same question, =(

To find the new volume of the neon gas sample when the pressure increases to 1.50 atm while the temperature remains constant, we can use the combined gas law.

The combined gas law relates the three variables: pressure (P), volume (V), and temperature (T) of a gas sample. It can be represented as:

(P1 * V1) / T1 = (P2 * V2) / T2

where P1, V1, and T1 denote the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 represent the final pressure, volume, and temperature, respectively.

In this case, we are given:
P1 = 0.950 atm (initial pressure)
V1 = 3.50 L (initial volume)
T1 = 20 degrees C (initial temperature)
P2 = 1.50 atm (final pressure)
T2 = 20 degrees C (final temperature)

Now, let's plug these values into the equation and solve for V2 (final volume):

(0.950 atm * 3.50 L) / (20 + 273.15 K) = (1.50 atm * V2) / (20 + 273.15 K)

To simplify the equation, we convert the temperature from degrees Celsius to Kelvin (K). The conversion formula is K = °C + 273.15.

Simplifying further, we get:

3.33 L / 293.15 K = 1.50 atm * V2 / 293.15 K

Cross-multiplying gives us:

3.33 L * 1.50 atm = V2 * 1.50 atm

4.995 L atm = 1.50 V2

Dividing both sides by 1.50 atm:

V2 = 4.995 L / 1.50 atm

V2 ≈ 3.33 L

Therefore, when the pressure increases to 1.50 atm while the temperature remains constant, the volume of the neon gas sample would be approximately 3.33 liters.

1. An ideal gas initially at 0.950 atm and 53oC is heated in a constant volume, sealed container to 100oC. What is the final pressure?

P1V1 = P2V2