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2.2 g of nitrous oxide \left(\mathrm{N}_{2} \mathrm{O}\right) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, \mathrm{\Delta \mathrm{U}} is \mathrm{'-x^{\prime} J}. The value of 'x' is _____.[nearest integer]

\text { (Given: atomic mass of } \mathrm{N}=14 \mathrm{~g} \mathrm{~mol}^{-1} \text { and of } \mathrm{O}=16 \mathrm{~g} \mathrm{~mol}{ }^{-1} \text {. }

\text { Molar heat capacity of } \mathrm{N}_{2} \mathrm{O} \text { is } 100 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text { ) }

Option: 1

195


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\text { moles of } \mathrm{N}_{2} \mathrm{O}=\frac{2.2}{44}=0.05}

\mathrm{ \Delta H=n C_{p} \Delta T =0.05 \times 100(270-310) }\\

                             \mathrm{=0.05 \times 100(-40)=-200 \mathrm{~J}}

\mathrm{W =-P_{e x t} \Delta V} \\

    \mathrm{=-1 \times\left(\frac{167.75-217.1}{1000}\right) \times 101.3 \mathrm{~J}} \\

\mathrm{W =+5 \mathrm{~J}}

\mathrm{\Delta U=q_{p}+W}\\

\mathrm{\Delta U=-200+5=-195 \mathrm{~J}}

\mathrm{value \: of\: ' x ' \: is \: 195}

Hence the answer is 195

Posted by

rishi.raj

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