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4 years ago

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An ideal gas initially at 275 k undergoes an isobaric expansion at 2.50 kpa. the volume increases from 1.00 m3 to 3.00 m3 and 11

.2 kj is transferred to the gas by heat. (a) what is the change in internal energy of the gas?

1 answer:

tekilochka [14]4 years ago

5 0

The transformation is isobaric (constant pressure), so the work done by the gas is the product between its pressure and the variation of volume:
$W=p \Delta V=(2500 Pa)(3.0 m^3-1.0 m^2 ) = +5000 J =+5.0 kJ$

The heat transferred to the gas is
$Q=+11.2 kJ$

So we can use the first law of thermodynamics to compute the variation of internal energy of the gas:
$\Delta U = Q-W=11.2 kJ-5.0 kJ=+6.2 kJ = +6200 J$
where the positive sign means the internal energy of the gas has increased.

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Answer:

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Explanation:

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$\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2$

(a)

The final angular speed is given as;

$\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s$

(b) the time taken to turn 5.5 revolutions is given as

$\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s$

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Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

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