At a given temperature the pressure of a gas reduces to $75$ of its initial value and the volume increases by $40$ of its initial value. Find this temperature if the initial temperature was $- 10^{o} C$ .

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Solution

$T_{1} = - 10^{0} C = 263 K$
$P_{1} = x$ $V_{1} = Y$
$T_{2} = ?$
$P_{2} = \frac{75 x}{100}$

$= \frac{75 x}{100}$

$V_{2} = \frac{140 y}{100}$

$= \frac{140 y}{100}$

$\frac{V_{1} P_{1}}{T_{1}} = \frac{V_{2} P_{2}}{T_{2}}$

$x y \times 263 = \frac{75 x \times 140 y}{10000 \times T_{2}}$

$\Rightarrow \frac{263 \times 10000}{75 \times 140} = T_{2}$

$\Rightarrow 250.47 = T_{2}$

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At a given temperature the pressure of a gas reduces to 75 of its initial value and the volume increases by 40 of its initial value. Find this temperature if the initial temperature was −10oC.

T1=−100C=263K P1=x V1=Y T2=? P2=75x100 =75x100 V2=140y100=140y100 V1P1T1=V2P2T2 xy×263=75x×140y10000×T2 ⇒263×1000075×140=T2⇒250.47=T2

At a given temperature the pressure of a gas reduces to $75$ of its initial value and the volume increases by $40$ of its initial value. Find this temperature if the initial temperature was $- 10^{o} C$ .