Question on: JAMB Physics - 2019

The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm\(^3\). Calculate the volume

A

2000.25 cm\(^3\)

B

200.025 cm\(^3\)

C

4003.27 cm\(^3\)

D

2010.25 cm\(^3\)

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Correct Option: B

m = 0.354g,

T\(_1\) = 273°C = 273 + 273 = 576K

P\(_1\) = 114cmHg, V\(_1\) = 2667cm\(^3\) at STP

T\(^2\) = 273K, P\(_2\) = 76cmHg, V\(_2\) = ?

But,

P\(^1\)V\(^1\)
T\(^1\)
= P\(^2\)V\(^2\)
T\(^1\)

 

V\(_2\) = 114 × 2667 × 273
76 × 576
= 2000.25cm\(^3\)

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