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Question

$$32$$ gram of oxygen gas at temperature $$27^o C$$ is compressed adiabatically to $$\dfrac{1}{3}$$ of its initial volume. Calculate the change in internal energy. ($$\gamma \ =\ 1.5$$ for oxygen)


Solution

For , adiabatically ,
$$dU = nCvdT =\dfrac {nR}{(\gamma-1)}dT$$
now, 
$$PV^{\gamma} = constant $$
so, $$V^{\gamma}\times\dfrac {nRT}V =constant $$
$$TV^{(\gamma -1)} = constant $$
so, $$T_1 V_1^{(\gamma-1)} = T_2V_2^{(\gamma-1)} $$
$$300 \times V^{(\gamma-1)} = T_2\times\left(\dfrac V3\right)^{(\gamma-1)} $$
$$\dfrac {300}{T_2} = \left(\dfrac 13\right)^{(1.5 -1)} $$
$$T_2 = \dfrac {300}{(1/\sqrt 3)} = 300\sqrt 3 K $$
n = no of mole = $$\dfrac {32}{32} = 1$$
now, 
$$dU = 1 \times \dfrac R{(1.5-1)} \times (300\sqrt 3 -300)$$
$$\implies = 2R \times  300( \sqrt 3 -1) $$
$$\implies =2 \times \dfrac {25}3 \times 300(1.732-1) $$
$$=50 \times 100 \times 0.732 J=3660J $$

Chemistry

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