Question

The volume at s.t.p. occupied by a gas 'Z' originally occupying 1.57 dm^3 at 310.5 K and 75 cm. press. of Hg.

Answer 1

Given: Pressure, P1 = 75 cm Hg = 750 mm Hg = 1 bar

Volume, V1 = 1.57 dm3 , Temperature, T1 = 310.5 K

At S.T.P. : Pressure, P2 = 1 bar ;  Temperature, T2 = 273.15 K ;   Volume, V2 = ?

P1V1T1=P2V2T21 bar×1.57 dm3310.5 K=1 bar×V2273.15 KV2=1 bar×1.57 dm3×273.15 K310.5 K ×1 bar=1.38 dm3

Hence, the volume at S.T.P. = 1.38 dm3

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Answer 2

Answer:

initial conditions

V1 is equal to 1.57dm3

P1=75cm

T1=310.5 K

final conditions

V2=?

P2=76 cm

T2 =273 K

using the gas equation P1V1/T1= P2V2/T2

therefore V2=P1 V1 T2 /T1 P2

= 75×1.57× 273/ 310.5 × 76

= 1.36dm3

Explanation:

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