The pressure P and absolute temperature T of an ideal gas are related to each other as P -1T2- If temperature of gas is 300 K and its volume changes from V to 8 V- then find the final temperature of gas

The correct option is A 600 K Solution:Pressure ,P α1T^2 (Given) ⇒ P= kT^2 We know, from the ideal gas equation,PV= nRT ⇒kT^2 #215; V =nRT ...

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Standard XII

Chemistry

Boyle's Law

The pressure ...

Question

The pressure (P) and absolute temperature (T) of an ideal gas are related to each other as $P \propto \frac{1}{T^{2}}$ . If temperature of gas is 300 K and its volume changes from V to 8 V, then find the final temperature of gas

600 K

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300 K

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373 K

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273 K

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Solution

The correct option is A $600 K$
Solution:
Pressure ,P $α \frac{1}{T^{2}}$ (Given)
$\Rightarrow P = \frac{k}{T^{2}}$
We know, from the ideal gas equation,
PV= nRT
$\Rightarrow \frac{k}{T^{2}} \times V = n R T$
$\Rightarrow V = \frac{n R T^{3}}{k}$
$\Rightarrow V α T^{3}$
Therefore, $\frac{V_{1}}{V_{2}} = \frac{T_{1}^{3}}{T_{2}^{3}}$
$\Rightarrow \frac{V}{8 V} = \frac{300^{3}}{T_{2}^{3}}$
$\Rightarrow T_{2}^{3} = 300 \times 300 \times 300 \times 8$
$\Rightarrow T_{2} = (300 \times 300 \times 300 \times 8)^{1 / 3}$
Therefore, $T_{2} = 300 \times 2$ = 600 K
Hence, option A is correct.

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The pressure (P) and absolute temperature (T) of an ideal gas are related to each other as P∝1T2. If temperature of gas is 300 K and its volume changes from V to 8 V, then find the final temperature of gas

The correct option is A 600 KSolution:Pressure ,P α1T2(Given)⇒P=kT2We know, from the ideal gas equation,PV= nRT⇒kT2×V=nRT⇒V=nRT3k⇒VαT3Therefore, V1V2=T31T32⇒V8V=3003T32⇒T32=300×300×300×8⇒T2=(300×300×300×8)1/3Therefore, T2=300×2= 600 KHence, option A is correct.

The pressure (P) and absolute temperature (T) of an ideal gas are related to each other as $P \propto \frac{1}{T^{2}}$ . If temperature of gas is 300 K and its volume changes from V to 8 V, then find the final temperature of gas