Consider the following two-step process. Heat is allowed to flow out of an ideal
ID: 2251784 • Letter: C
Question
Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 0.6 atm to 0.2 atm. Then the gas expands at constant pressure, from an initial volume of 208 L to a final volume at which the temperature reaches its original value. See the illustration at right. Calculate the total heat flow, to the nearest joule, into or out of the gas.
Explanation / Answer
for the 1st process P1 * V1 = n * R * T1.... (1)
2nd process P2 * V1 = n * R * T2..... (2)
3rd process P2 * V2 = n * R * T1 ...... (3)
so (3) / ( 1)
V2 = P1 / P2 * V1 = 3 * 208 = 624 L
Q = m * C v ( T2 - T1) + m * C p ( T1 - T2 ) = m * ( T2 - T1 ) * ( Cv - C p )
C p - Cv = R
so Q = m * ( T1 - T2 ) * R
= m * R * T1 - m* R * T2 = P1 * V1 - P2 * V2
P1 = 0.6 atm = 60794.998598 N / m^2
P2 = 0.2 atm = 20264.999533 N/ m^2
V1 = 208 L = 208 * 10^-3 m^3
V2 = 624 L = 624 * 10^-3 m^3
so Q = 4.16 * 10^-3 J