2P 2 V 2003 Thomson Brooks/Cole A quantity of a monatomic ideal gas undergoes a

Question

2P 2 V 2003 Thomson Brooks/Cole A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are doubled as shown n the figure above. DATA: Vo- 0.31 m Po 9500 Pa What is the change of the internal energy of the gas? Submit Answer Tries 0/20 What was the work done by the gas during the expansion? submit Answer Incompatible units. No conversion found between "w" and the required units Tries o/20 Previous Tries What amount of heat flowed into the gas during the expansion?

Explanation / Answer

?U = Q - W
Q = ?U + W

use ideal gas law:
p0V0 = nRT0 and
p1V1 = nRT1
---> T1 = p1V1/(nR)
with p1 = 2p0 and V2 = 2V1
T1 = 2p0*2V0/(nR) = 4p0V0/(nR)

?T = T1 - T0 = 4p0V0/(nR) - p0V0/(nR) = 3p0V0/nR
And ?U = Cv*?T with Cv for an ideal gas = 3/2*R
So ?U = n*3/2*R*3p0V0/(nR) = 4.5p0V0
?U = 4.5*9500*0.31 J = 13252.5 J (change of internal energy)

Work done is
W = 1/2(p0+p1)(V1-V0) = 1/2(p0+2p0)(2V0-V0)
= 1/2(3p0)(V0) = 1.5p0V0
W = 1.5*9500*0.31 = 4417.5 J (work done)

Q = ?U + W = 13252.5+4417.5 = 17670 J (heat flow into gas)

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