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:
<span>12 (2T1 + 273.15) = 2 P2 (T1 + 273.15)</span>
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)
<span>P2 = 6.312 atm</span>
