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A cylinder of nitrogen gas has a volume of 15,000cm3and a pressure of100 atm.
a. What is the thermal energy of this gas at room temperature(20C)?
b. What is the mean free path in the gas?
c. The valve is opened and the gas is allowed to expand slowly and isothermally until it reaches a pressure of 1.0atm . What is the change in the thermal energy of the gas?

Short Answer

Expert verified

a)The Thermal energy of gas isE=379KJ.
b)The Mean free path in tha gas isλ=2.25nm.
c)The change in thermal energy of the gas is Zero.

Step by step solution

01

Step: 1   Finding the Thermal Energy of gas at  room temperature(20∘C): (part a)

Because nitrogen is a diatomic gas, the thermal energy is calculated as follows:

E=52NKBT

Now we know that , pV=NKBT

From equation localid="1648381889785" 1and2we get that

E=52pV

Given thatp=100atm=1.01×107N/m2and V=15000cm3=0.015m3,, then substituting these to the formula:

localid="1648390595486" E=52×1.01×107×0.015=379kJ.

02

Step: 2  Finding the Mean free path: (part b)

The mean free path is given by

λ=142π(N/V)r2

Now we have

NV=pKBT=1.01×1071.38×1023×293=2.5×1027

and rfor nitrogen is r=1×1010,then the mean free path will be:

λ=142π×1027×1×10102=2.25×109m=2.25nm.

03

Step: 3  Change in Thermal Energy of gas: (part c)

But even though the formula shows that thermal energy is proportional to the number of gas molecules, degrees of freedom, and gas temperature, none of these factors change in this isothermal expansion.

As a result, the total amount of thermal energy will remain unchanged. Thermal energy will remain unchanged.

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