We have a container of 1.85 moles of an ideal monatomic gas. The volume of the c
ID: 1471256 • Letter: W
Question
We have a container of 1.85 moles of an ideal monatomic gas. The volume of the container is 15.0 liters, and the temperature of the gas is 21.7C. We compress the gas adiabatically to 12.9 liters. (a) Find the final temperature (K) of the gas. Neglect any heat flow into the surroundings. Caution: Be sure to use the ideal value for (the fraction,) not the approximate (decimal) value. (b) Find the change in internal energy (J) of the gas. (c) Find the work done (J) on the gas. Be sure to include the correct signs on the answers.
Explanation / Answer
= 5/3 for monoatmoic gas
Isentropic process obeys the following equation:
PV^ = constant, or
P1V1^ = P1V2^
We know for ideal gas,
PV = nRT
P = nRT/V,
T2/T1 = (V1/V2)^(-1)
T2 = 294.86 *(15/12.9)^(5/3 -1)
T2 = 326.1 K
(b)
PV = nRT
P = nRT/V
P = (1.85 * 8.314 * 294.86)/(0.015)
P1 = 3.02 * 10^5 Pa
pv^y = k = 277.74
The definition of an adiabatic process is that heat transfer to the system = 0
Therefore,
Change in Internal Energy + Work done = 0
In Compression, Work is done on the Gas, Therefore work done is Negative.
Work done in adiabatic process is given by,
Work done = k * vf^(1-y) - vi^(1-y)/(1-y)
Work done = 277.74 * (0.0129^(1- 5/3) - 0.015^(1-5/3)) / (1 - 5/3)
Work done = -724.5 J
(b)
Internal Energy = 724.5 J
(c)
Work done = - 724.5 J
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