# PHYSICS

Suppose that a tank contains 680m^3 of neon at an absolute pressure of 1.01*10^5 Pa. The temperature is changed from 293.2 to 294.3 K. What is the increase in the internal energy of the neon?

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1. The change in internal energy is
delta U = (3/2)R*(delta T)*n
where n is the number of moles.

delta T is only 1.1 K in this case.

Note: (3/2)R is the specific heat at constant volume of any monatomic gas

Get the number of moles from
n = PV/RT
Use the initial P and T.
V = 680 m^3 = 680*10^3 liters. P is 1.00 atm.
I get 2.83*10^4 moles for n.

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2. he change in internal energy of a gas (or any fluid) is given by:
?U = n*(u2 - u1)

If you assume the neon is calorically perfect, then this can be alternatively written as:
?U = n*cv*(T2 - T1)

where n is the number of moles of gas, cv is the molar isochoric specific heat of the gas, and T2 and T1 are the final and initial temperatures.

cv can be expressed in terms of the gas constant R (independant of gas flavor), and k, the adiabatic index, specific to the gas flavor. For all monatomic gasses (He, Ne, Ar, Kr, Xe), or even hydrogen when it exists monatomically on the sun, k=5/3.
cv = R/(k - 1)

Re-construct:
?U = n*R/(k - 1)*(T2 - T1)

Use the ideal gas law to find n*R
P1*V1 = n*R*T1

solve for n*R:
n*R = P1*V1/T1

Substitute:
?U = P1*V1*(T2/T1 - 1)/(k - 1)

Data:
P1:=101 kPa; V1:=680 m^3; T1:=293.2 K; T2:=294.3 K; k:=5/3;

do realize that 1 kPa * 1m^3 equals 1 kiloJoule.

Result:
?U = 386.5 kiloJoules

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