Question

☑100 points☑

give full explanation...​

Answer 1

❤❤.. $\huge\mathfrak{Hlo\:Reetu}$... ❤❤

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W=1000kg

∆T=60°C

No. Of moles of air=1000/0.029

n=34483

Now,

∆E=nR∆T

∆E=34482×8.314×60

∆E=17.2×10^6. J

Answer 2

Given

W= 1000 kg

AT 60 C =60 K

Solution

-as standard convention, molar mass of air is taken as 0.029 kg / mol

No of moles of air in balloon

-n 1000 / 0.029

n 34483

Change in thermal energy of air is calculated by

DeltaE=nRdeltaT

DetlaE=34482 x 8.314 x 60

DeltaE =17.2x10^ 6J

So, change in thermal energy of 17.2x10^ 6 J.