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

Answer:

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

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:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (Constant pressure)

$T_2=\frac{284.15 K\times 3V}{V}=852.45K=579 ^oC$

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