1. Tardigrade
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  3. Chemistry
  4. Two mole of an ideal gas is heated at constant pressure of one atmosphere from 27° C to 127° C. If C v , m =20+10-2 T JK -1 mol -1, then q and Δ U for the process are respectively:

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Q. Two mole of an ideal gas is heated at constant pressure of one atmosphere from $27^{\circ} C$ to $127^{\circ} C$. If $C_{ v , m }=20+10^{-2} T \,JK ^{-1} mol ^{-1},$ then $q$ and $\Delta U$ for the process are respectively:

Thermodynamics

Solution:

$w =- nR \,\Delta T =-2 \times 8.314 \times 100=-1662.8 \,J$

$\Delta U = n$

$ \int \limits_{300}^{400}\left(20+10^{-2} T\right) d T$

$=2\left[2 \times 100+\frac{10^{-2}}{2}\left(400^{2}-300^{2}\right)\right]=4700 \,J$

$4700= q -1662.8$

$\therefore q =6362.8 \,J$