Question 1

Multiple choiceHard1 mark

A sample of 6.0×10236.0 \times 10^{23} particles of a monatomic gas is initially at 400 K400\ \mathrm{K} In a final state, its volume is twice its initial volume, its number density is 3/43/4 of the initial value, and its temperature is 2/32/3 of the initial value. At both states the gas is sufficiently hot and dilute for the ideal-gas approximation to apply. What is its final internal energy? Take kB=1.38×10−23 J K−1k_B=1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}

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Answer: C. The particle-number ratio is the number-density ratio multiplied by the volume ratio: Nf/Ni=(3/4)(2)=3/2N_f/N_i=(3/4)(2)=3/2 Since U=32NkBTU=\frac{3}{2}Nk_BT the internal-energy ratio is (3/2)(2/3)=1(3/2)(2/3)=1 Thus Uf=Ui=32(6.0×1023)(1.38×10−23)(400)=4.968×103 JU_f=U_i=\frac{3}{2}(6.0\times10^{23})(1.38\times10^{-23})(400)=4.968\times10^3\ \mathrm{J} or 5.0×103 J5.0\times10^3\ \mathrm{J} to two significant figures.
A: treats the number-density ratio as the particle-number ratio, omitting the volume change.
B: accounts for the temperature change but holds the particle number fixed.
D: treats the volume ratio as the particle-number ratio, ignoring the number-density change.
Remember: for an ideal monatomic gas, internal energy depends on particle number and absolute temperature.

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