What work was done on two moles of an ideal monatomic gas during its adiabatic compression, if its temperature increased by 20 K.
| November 22, 2020 | education
Given
v = 2 mol
i = 3 (degree of freedom)
dT = 20 K
R = 8.31 – gas constant
adiabatic
find: A
Decision:
in the adiabatic process Q = 0
work is equal to the change in internal energy A = dU
for a monatomic gas dU = i / 2 vR * dT = 3/2 vR * dT = 3/2 * 2 * 8.31 * 20 = 498.6 J
answer
A = 498.6 J
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