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Question: When one mole of monatomic ideal gas at T K undergoes adiabatic change under a constant external pre...

When one mole of monatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 liter to 2 liters. The final temperature in Kelvin would be?
(A)T223\dfrac{T}{{{2}^{\dfrac{2}{3}}}}
(B)T+23×0.0821T+\dfrac{2}{3\times 0.0821}
(C) T
(D) T−23×0.0821T-\dfrac{2}{3\times 0.0821}

Explanation

Solution

As we know that at adiabatic conditions, q=0 and hence by first law of thermodynamics:-
ΔU=ΔW\Delta U=\Delta W. Also the value of ratio of specific heats(γ)(\gamma ) for monatomic gas is 5/3.

Formula used: We will require the following formulas:-
ΔU=nCV(T2−T1)\Delta U=n{{C}_{V}}({{T}_{2}}-{{T}_{1}})
ΔW=−Pext(V2−V1)\Delta W=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})
CP−CV=R{{C}_{P}}-{{C}_{V}}=R

Complete step-by-step answer:
According to the First Law of Thermodynamics:-
ΔU=ΔW+q\Delta U=\Delta W+q
But since we are given adiabatic condition which means q=0. Therefore, ΔU=ΔW\Delta U=\Delta W.
As we know that,
ΔU=nCV(T2−T1)\Delta U=n{{C}_{V}}({{T}_{2}}-{{T}_{1}})
ΔW=−Pext(V2−V1)\Delta W=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})
-Therefore, putting them inΔU=ΔW\Delta U=\Delta W, we get:-
nCV(T2−T1)=−Pext(V2−V1)−−−eq(1)n{{C}_{V}}({{T}_{2}}-{{T}_{1}})=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})---eq(1)

-We have been given the following values:-
V1{{V}_{1}} =1L
V2{{V}_{2}}= 2L
T1{{T}_{1}} = T K
Pext{{P}_{ext}}= 1atm
R = universal gas constant = 0.0821atm⋅Lmol⋅K0.0821\dfrac{atm\cdot L}{mol\cdot K}
Also the value of ratio of specific heats for monatomic gas is 5/3.

-By usingCP−CV=R{{C}_{P}}-{{C}_{V}}=R, we get:-

⇒CP−CV=R\Rightarrow {{C}_{P}}-{{C}_{V}}=R
divide whole equation by CV:−\text{divide whole equation by }{{C}_{V}}:-
⇒CPCV−1=RCV\Rightarrow \dfrac{{{C}_{P}}}{{{C}_{V}}}-1=\dfrac{R}{{{C}_{V}}}
⇒γ−1=RCV ⇒CV=Rγ−1 ⇒CV=R53−1 ⇒CV=R23  \Rightarrow \gamma -1=\dfrac{R}{{{C}_{V}}} \\\ \Rightarrow {{C}_{V}}=\dfrac{R}{\gamma -1} \\\ \Rightarrow {{C}_{V}}=\dfrac{R}{\dfrac{5}{3}-1} \\\ \Rightarrow {{C}_{V}}=\dfrac{R}{\dfrac{2}{3}} \\\
On substituting all these values in eq(1), we get:-
⇒1×0.08212/3(T2−T)=−1(2−1) ⇒1×0.08212/3(T2−T)=−1 ⇒(T2−T)=−1×2/30.0821 ⇒T2=T−23×0.0821  \Rightarrow 1\times \dfrac{0.0821}{2/3}({{T}_{2}}-T)=-1(2-1) \\\ \Rightarrow 1\times \dfrac{0.0821}{2/3}({{T}_{2}}-T)=-1 \\\ \Rightarrow ({{T}_{2}}-T)=-1\times \dfrac{2/3}{0.0821} \\\ \Rightarrow {{T}_{2}}=T-\dfrac{2}{3\times 0.0821} \\\
Hence the final temperature is:-
Option (D) T−23×0.0821T-\dfrac{2}{3\times 0.0821}.

Note:
- Always write down the given values provided in the question and if required, change their values accordingly so as to bring them into the same units as that of the universal gas constant.
- The ratio of specific heats(γ)(\gamma ) for monatomic gas is 5/3 and diatomic gas is 7/5.