Question

a 3.0 l container of neon gas has a pressure of 1.0 atm. what will the pressure be if the volume is decreased to 1.3 l at a constant temperature?

Answer 1

2.31 atm

Step-by-step explanation:

Use the combined gas law:

P1V1/T1 = P2V2/T2

where P, V, and T are pressure (P), volume (V), and temperature (T). P1, V1, and T1 are the initial states and P2, V2, and T2 are the final states. Temperature must be in Kelvin.

We are given:

V1 = 3.0L

P1 = 1.0 atm

V2 = 1.3, and

temperature is unknown, but held constant: T2 = T1

Rearrange P1V1/T1 = P2V2/T2 for the unknown, P2:

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

Enter the data:

P2 = (1.0atm)((3.0L)/(1.3L))(T1/T1) [The temperatures cancel]

P2 = (1.0atm)((3.0L)/(1.3L)

P2 = (1.0atm)(2.31)

P2 = 2.31 atm