Question

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?

Answer 1

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