Question

Question: A Sample Consisting Of 1 Mol Of Dioxygen Molecules With Cp,M= 29.4 JK-1mol-1 Has An Initial Volume Of 0.05m3 At T1 = 270K. It undergoes reversible adiabatic compression until its volume reaches 0.02m3. calculate the work for this process

Answer 1

Work for the reversible Adiabatic compression process is 2.5118 KJ

Step by step explanation:

Given

V1 (initial volume)=0.05 $m^{3\\}$  

V2(final volume) = 0.02 $m^{3\\}$

T1 (initial temperature) = 270 K

n(no. of moles) =$1 mole$

Cp,M = 29.4 J$K^{-1}mol^{-1}$  

For solving this question we need γ(degree of freedom),  Cv, M, and final temperature(T2).

Step 1 - Calculation for T2(final temperature);

we will use, $T1(V1)^{\gamma-1}$ = $T2(V2)^{\gamma-1}$  OR   $\frac{T1}{T2}$ = $(\frac{V2}{V1} )^{\gamma-1}$    

as we have a diatomic (O₂) molecule and we know γ of a diatomic molecule is 1.4, by using this γ we get our final temperature T2 which is 389.61 K

Step 2 - Calculation for Cv, M;

As we know   = γ, by this we can find Cv, M, after calculation we get Cv, M = $21 J$ $K^{-1}mol^{-1}$

Step 3 - Calculation of W(work done);

now we have all the required things with us for the calculation of Work done;

W=n (Cv, M) (T2-T1) by putting all the values, we will get work done

(W)= $2511.82J$ and we can convert it into KJ and the W will be $2.5118 KJ$.