We have a container of 1.98 moles of an ideal monatomic gas. The volume of the c

Question

We have a container of 1.98 moles of an ideal monatomic gas. The volume of the container is 15.0 liters, and the temperature of the gas is 21.7C. We compress the gas adiabatically to 13.7 liters.

(a) Find the final temperature (K) of the gas. Neglect any heat flow into the surroundings. Caution: Be sure to use the ideal value for (the fraction,) not the approximate (decimal) value.

(b) Find the change in internal energy (J) of the gas.

(c) Find the work done (J) on the gas. Be sure to include the correct signs on the answers.

THank you:)

Explanation / Answer

initial volume=v1=15 ltrs

initial temperature=21.7 degree celcius=21.7+273=294.7 K

final volume=13.7 ltrs

number of moles=n=1.98

value of gamma for monoatomic gas is 5/3.

as we know, for adiabatic process, P*V^(gamma)=constant

==>(n*R*T/V)*V^(gamma)=constant

as n and R are constant, T*V^(gamma-1)=constant

using gamma=5/3

T*V^(2/3)=constant

==>294.7*15^(2/3)=T*13.7^(2/3)

==>T*13.7^(2/3)=1792.4

==>T=1792.4/(13.7^(2/3))=313.06 K

hence final temperature of the gas is 313.06 K or 40.055 degree celcius.

b)

change in internal energy=n*Cv*change in temperature

for monoatomic gas, Cv=(3/2)*R

hence change in internal energy=1.98*1.5*8.314*(313.06-294.7)=453.36 J

c)let work done on the gas is W.

then as we know, in adiabatic process, total heat transferred=0

hence work done on the gas=total change in internal energy of the gas (from first law of thermodynamics)

hence work done on the gas=453.36 J

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