Question

A 22.8 L tank of Ar gas is connected to an evacuated 9.26 L tank. If the final pressure is

40.8 mmHg, what must have been the original gas pressure in the tank?

Answer 1

57.4 mmHg.

Step-by-step explanation:

We can use the ideal gas law to solve this problem:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

Since the two tanks are connected, we can assume that the number of moles of gas is conserved. Therefore, we can write:

P1V1 = P2V2

where:

P1 = initial pressure (what we want to find)

V1 = initial volume (22.8 L)

P2 = final pressure (40.8 mmHg)

V2 = final volume (22.8 L + 9.26 L = 32.06 L)

Rearranging the equation, we get:

P1 = (P2V2)/V1

Substituting the given values, we get:

P1 = (40.8 mmHg x 32.06 L) / 22.8 L

P1 = 57.4 mmHg

Therefore, the original gas pressure in the tank was 57.4 mmHg.