$2500 c m^{3}$ of hydrogen at S.T.P. is subjected to a $\frac{1}{2}$ times increase in pressure (temperature remaining constant). What volume will the hydrogen now occupy?

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Solution

$P_{1} V_{1} = P_{2} V_{2}$
Here, $P_{1} = 1 a t m, V_{1} = 2500 c m^{3}, P_{2} = 1 + 1 \times 5 / 2 = 7 / 2 a t m$ ,
Then, $V_{2} = \frac{1 \times 2500}{\frac{7}{2}} = 5000 / 7 c m^{3}$

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$2500 c m^{3}$ 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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2500cm3 of hydrogen at S.T.P. is subjected to a 12 times increase in pressure (temperature remaining constant). What volume will the hydrogen now occupy?

P1V1=P2V2Here, P1=1 atm, V1=2500 cm3, P2=1+1×5/2=7/2 atm,Then, V2=1×250072=5000/7 cm3

$2500 c m^{3}$ of hydrogen at S.T.P. is subjected to a $\frac{1}{2}$ times increase in pressure (temperature remaining constant). What volume will the hydrogen now occupy?

$P_{1} V_{1} = P_{2} V_{2}$
Here, $P_{1} = 1 a t m, V_{1} = 2500 c m^{3}, P_{2} = 1 + 1 \times 5 / 2 = 7 / 2 a t m$ ,
Then, $V_{2} = \frac{1 \times 2500}{\frac{7}{2}} = 5000 / 7 c m^{3}$