$2500 c m^{3}$ of hydrogen is taken at S.T.P. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume will hydrogen occupy now?

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Solution

$P_{1} = 1 a t m = 760 m m, P_{2} = 760 \times 2.5 + 760 = 2660 m m$
$T_{1} = T_{2} = 273 K [c o n s t a n t]$
$V_{1} = 2500 c m^{3}, V_{2} = ?$

By gas equation,
$\frac{P 1 V 1}{T 1}$ = $\frac{P 2 V 2}{T 2}$
=> $\frac{760 \times 2500}{273}$ = $\frac{2660 \times V_{2}}{273}$
So $V_{2} = 714.28 c m^{3}$

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2500cm3 of hydrogen is taken at S.T.P. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume will hydrogen occupy now?

P1=1atm=760mm,P2=760×2.5+760=2660mmT1=T2=273K[constant]V1=2500cm3,V2=? By gas equation, P1V1T1 = P2V2T2 => 760×2500273 = 2660×V2273So V2=714.28cm3

$2500 c m^{3}$ of hydrogen is taken at S.T.P. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume will hydrogen occupy now?