Question

9. If the volume of a gas collected at a temperature
of 600 °C and pressure of 1.05 * 105 Nm? is 60
dms, what would be the volume of gas at STP
(P=1.01 * 103 Nm?, T = 273 K)?
A) 25 cm
C) 100 cm
B) 75 cm
D) 51 cm​

Answer 1

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Answer 2

Explanation:

If the volume of a gas collected at a temperature of 600 °C and pressure of 1.05 × 10⁵ Nmˉ² is 60 dm³, then what would be the volume of gas at STP (P=1.01 × 10³ Nmˉ², T=273 K)?

(note: STP pressure is 1.01 × 10⁵ Nmˉ², not 1.01 × 10³ Nmˉ².)

Use PV = nRT. Solving for V, we have V = nRT/P; n and R are constants here.

V₁ = 60 dm³.

P₁ = 1.05 × 10⁵ Nmˉ².

T₁ = 600 °C = 873 K.

V₂ = x dm³.

P₂ = 1.01 × 10⁵ Nmˉ².

T₂ = 273 K.

V2V1=(nRT2P2)(nRT1P1).V2V1=(nRT2P2)(nRT1P1).

Simplifying and solving for V₂, we get

V2=V1P1T2P2T1.V2=V1P1T2P2T1.

Substitute the given numbers and get out your calculator