Question

A sample of argon gas occupies a certain volume at 11°C. At what temperature would the volume of the gas be three times as big? Assume the pressure remains constant. 33.0°C 579°C 852°C 109°C

Answer 1

V₁/T₁ = V₂/T₂ = const

T₁=11+273=284K
V₂=3V₁

V₁/T₁=3V₁/T₂

T₂=3T₁

T₂=3*284=852K

t₂=852-273=579°C

Answer 2

At 579°C the volume of the gas be three times as big.

Step-by-step explanation:

Under constant pressure:

Initial temperature of the gas =
`T_1=11^oC=284.15K`

Initial volume of the gas =
`V_1=V`

Initial temperature of the gas =
`T_2=?`

Initial volume of the gas =
`V_2=3V`

Applying Charles law:

`(V_1)/(T_1)=(V_2)/(T_2)` (Constant pressure)

`T_2=(284.15 K* 3V)/(V)=852.45K=579 ^oC`

At 579°C the volume of the gas be three times as big.