1.) How much work is done on the external environment by 7 moles of a monatomic
ID: 1570903 • Letter: 1
Question
1.) How much work is done on the external environment by 7 moles of a monatomic ideal gas initially at 20oC that is ISOBARICALLY expanded at 1 atm to 5 times its original volume?
a. 2.56 x 104 J
b. 0 J
c. 3.41 x 104 J
d. 6.82 x 104 J
2.) What is the final temperature of the gas (in Celsius)?
a. 3.46 x 103 oC
b. 2.03 x 103 oC
c. 2.74 x 103 oC
d. 1.19 x 103 oC
3.) What is the change in internal energy of the gas?
a. 2.56 x 104 J
b. 1.02 x 105 J
c. 6.82 x 104 J
d. 1.28 x 105 J
4.) How much heat is absorbed by the gas during the expansion? Calculate this in two ways: First, use your answers above for the change in internal energy and the work done. Secondly, use the specific heat at constant pressure and the change in temperature. Compare your results (to 3 sig figs)
a. -1.02 x 105 J
b. 6.82 x 104 J
c. 2.56 x 104 J
d. 1.71 x 105 J
Explanation / Answer
(1)
From ideal gas law,
PV = nRT
V = 7*8.314*293 / 1.013*10^5 = 0.16829
Work done by gas,
W = -P*dV = 1.013*10^5*(5V - V)
W = -6.82*10^4 J
(2)
at constant pressure,
V1 / V2 = T1 / T2
V / 5V = 293 / T2
T2 = 1465 k = 1.19*10^3 oC
(3)
change in internal energy,
dU = n*cv*dT
For monoatomic gas, cv = 3R / 2
dU = 7 * 3*8.314 / 2 * (1192 - 20)
dU = 1.02*10^5 J
(4)
From first law of thermodynamics
Q = dU - W
Q = 1.02*10^5 - (-6.82*10^4)
Q = 1.71*10^5 J
We know that,
Q = n*cp*dT
Q = 7 * 5*8.314 / 2 * (1192 - 20)
Q = 1.71*10^5 J
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