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2 years ago

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An ideal gas at 78C is in a spherical flexible container having a radius of 1.00 cm. The gas is heated at constant pressure to 8

88C. Determine the radius of the spherical container after the gas is heated. [Volume of a sphere 5 (4y3)pr3 .]

1 answer:

Iteru [2.4K]2 years ago

8 0

Answer:

Explanation:

Given

initial temperature $T_1=78^{\circ}C\approx 351\ K$

initial radius $r_1=1\ cm$

final temperature $T_2=888^{\circ}C\approx 1161\ K$

As the gas heated at constant pressure

$P_1=P_2$

using $PV=nRT$

thus

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$\frac{\frac{4\pi r_1^3}{3}}{T_1}=\frac{\frac{4\pi r_2^3}{3}}{T_2}$

cancel out common terms

$\frac{r_1^3}{T_1}=\frac{r_2^3}{T_2}$

$(\frac{r_1}{r_2})^3=\frac{T_1}{T_2}$

$(\frac{1}{r_2})^3=\frac{351}{1161}$

$\frac{1}{r_2}=0.6711$

$r_2=1.489\ cm$

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The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of $\sqrt{3}$ , (d) a factor of 1, or (e) a factor of $\frac{1}{3}$ ?

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For 300K:

$V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }$

$V_{rms1}=30\sqrt{\frac{k_{B}}{m} }$

For 900K:

$V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }$

$V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }$

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$\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }$

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For rms, factor of change is $\sqrt{3}$

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q = m.v

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