Question
Calculate the fall in temperature of helium initially at 15°C when it is suddenly expanded to 8 times its original volume (? = 5/3).
✓
Answer
Given:
$ T _{ i }=15^{\circ} C =15+273=288 K$
$\gamma=\frac{5}{3}, V _{ f }=8 V _{ i } $
To find: Fall in temperature $(\Delta T)$
Formulae: $T _{ f } V _{ f }^{\Upsilon-1}= T _{ i } V _{ i }^{\Upsilon-1}$
Calculation:
From formula,
$ T _{ f }= T _{ i }\left(\frac{ V _{ i }}{ V _{ f }}\right)^{\Upsilon-1}$
$=288\left(\frac{1}{8}\right)^{\frac{5}{3}-1}$
$=288\left(\frac{1}{8}\right)^{\frac{2}{3}}$
$=288 \times \frac{1}{4}$
$=72 K$
$=72-273$
$=-201^{\circ} C$
$\therefore \Delta T = T _{ f }- T _{ i }$
$=-201-15 $
Fall in the temperature $(\Delta T )$ is $-216^{\circ} C$.
$ T _{ i }=15^{\circ} C =15+273=288 K$
$\gamma=\frac{5}{3}, V _{ f }=8 V _{ i } $
To find: Fall in temperature $(\Delta T)$
Formulae: $T _{ f } V _{ f }^{\Upsilon-1}= T _{ i } V _{ i }^{\Upsilon-1}$
Calculation:
From formula,
$ T _{ f }= T _{ i }\left(\frac{ V _{ i }}{ V _{ f }}\right)^{\Upsilon-1}$
$=288\left(\frac{1}{8}\right)^{\frac{5}{3}-1}$
$=288\left(\frac{1}{8}\right)^{\frac{2}{3}}$
$=288 \times \frac{1}{4}$
$=72 K$
$=72-273$
$=-201^{\circ} C$
$\therefore \Delta T = T _{ f }- T _{ i }$
$=-201-15 $
Fall in the temperature $(\Delta T )$ is $-216^{\circ} C$.
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