Question

If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the new temperature be to maintain constant pressure?

Answer 1

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$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$</p><p>\frac{15.0l}{298.0k} = \frac{45.0l}{x}298.0k15.0l=x45.0l \\ </p><p>\begin{gathered}15l \\ = \frac{45l}{x} \times 298.0k \\ \\ 15l = \\ \frac{13410lk}{x} \\ \\ 15lx = \\ 13410lk\end{gathered}15l=x45l×298.0k15l \\ =x13410lk15lx \\ =13410lk \\ </p><p>Cancel L \\ \begin{gathered}15x = 13410k \\ \\ \frac{15x}{13410} = k \\ \\ \frac{x}{894} = k\end{gathered} \\ 15x=13410k13410 \\ 15x=k894x=k</p><p>$

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Answer 2

Given :-

If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters

To Find :-

New temperature

Solution :-

We know that

0⁰ C = 273⁰ K

Volume/Temperature = Volume/Temperature

15/298 = 45/T

15 × T = 45 × 298

15T = 13,410

T = 13410/15

T = 894⁰ K

Now

0⁰ C = 273⁰ K

894⁰ K = 894 - 273 = 621⁰ C

$\begin{gathered} \\ \end{gathered}$