Question

During an experiment an ideal gas is found to obey an additional law vp2= constant. the gas is initially at a temperature 't' and volume 'v'. when it expands to a volume 3v, the temperature becomes

Answer 1

The answer for this question is:

Answer 2

$\sqrt{3} T$  

Explanation:

PV = RT...(1)

Given,

Ideal gas obeying additional law: $VP^{2}$ = constant....(2)

$P = \frac{RT}{V}$

$P^{2} = (\frac{RT}{V} )^{2}$

By equation (2),

$(\frac{RT}{V} )^{2} V = constant$ = $\frac{T_{1}^{2} }{3V_{1} }$ = $\frac{T^{2} }{V_{1} }$

⇒ $T_{1} ^{2}$ = $\frac{3V_{1} * T^{2} }{V_{1} }$

⇒ $T_{1}^{2}$ = $3T^{2}$

⇒ $T_{1} = \sqrt{3}T$

Learn more: law constant

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