Consider a sample of gas in a container on a comfortable spring day in chicago, il. the celsius temperature suddenly doubles, and you transfer the gas to a container with twice the volume of the first container. if the original pressure was 12 atm, what is a good estimate for the new pressure?

To solve this problem, we must first assume that the gas acts like an ideal gas so that we can use the ideal gas equation:

P V = n R T

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the absolute temperature

Assuming that the number of moles is constant, then we can write all the variables in the left side:

P V / T = k where k is a constant (n times R)

Equating two conditions or two states:

P1 V1 / T1 = P2 V2 / T2

We are given that V2 = 2 V1 therefore

P1 V1 T2 = P2 (2V1) T1

P1 T2 = 2 P2 T1

Additionally we are given that the temperature in Celsius is doubled, however in the formula we use the absolute temperature in Kelvin, therefore:

T1 (K) = T1 + 273.15

T2 (K) = 2T1 + 273.15

and P1 = 12 atm

Substituting:

12 (2T1 + 273.15) = 2 P2 (T1 + 273.15)

P2 = 6 (2T1 + 273.15) / (T1 + 273.15)

Assuming that a nice spring day in Chicago has a temperature of 15 Celsius, therefore:

P2 = 6 (2*15 + 273.15) / (15 + 273.15)

P2 = 6.312 atm