A 0.25 mol sample of neon gas at 18 °C and  0.799 atm is heated to 49 °C. The new gas pressure is 1.11 atm. What is the change in volume of the neon gas?

To solve this problem, we can use the combined gas law which states that the ratio of the product of pressure and volume to the temperature of a system remains constant. The formula for the combined gas law is:

P1V1/T1 = P2V2/T2

Where: P1 = initial pressure V1 = initial volume T1 = initial temperature P2 = final pressure V2 = final volume T2 = final temperature

Given in the problem: P1 = 0.799 atm T1 = 18 °C = 18 + 273.15 = 291.15 K (converted from Celsius to Kelvin) P2 = 1.11 atm T2 = 49 °C = 49 + 273.15 = 322.15 K (converted from Celsius to Kelvin)

We are asked to find the change in volume, which is V2 - V1.

First, let's solve for V2 using the combined gas law:

V2 = P1V1T2 / (P2*T1)

We don't know V1, but we know that the number of moles and the gas constant R are constant, so we can use the ideal gas law to express V1:

V1 = nRT1 / P1

Where: n = number of moles = 0.25 mol R = ideal gas constant = 0.0821 Latm/(Kmol)

Substituting V1 into the combined gas law gives:

V2 = P1*(nRT1 / P1)T2 / (P2T1) V2 = nRT2 / P2

Substituting the given values gives:

V2 = 0.25 mol * 0.0821 Latm/(Kmol) * 322.15 K / 1.11 atm V2 = 5.74 L

Similarly, we can find V1:

V1 = nRT1 / P1 V1 = 0.25 mol * 0.0821 Latm/(Kmol) * 291.15 K / 0.799 atm V1 = 7.57 L

Finally, the change in volume is V2 - V1 = 5.74 L - 7.57 L = -1.83 L.

So, the volume of the neon gas decreased by 1.83 L.