Question

Charle's Law

Answer 1

Convert temperature to Kelvin

• 225°C=498K
• 127°C=400K

Convert vol to L

• 400mL=0.4L

Apply Charles law

• V1T_2=V2T_1
• 0.4(400)=498V_2
• 160=498V_2
• V_2=0.32L=320mL

Answer 2

Answer:

Given:

$V_{1} = \text{400.0 mL}$

$T_{1} = 225.0^{\circ}\text{C + 273.15 = 498.15 K}$

$T_{2} = 127.0^{\circ}\text{C + 273.15 = 400.15 K}$

Unknown:

$V_{2}$

Solution:

$\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}$

$V_{2} = V_{1} \times \dfrac{T_{2}}{T_{1}}$

$V_{2} = \text{400.0 mL} \times \dfrac{\text{400.15 K}}{\text{498.15 K}}$

$\boxed{V_{2} = \text{321.3 mL}}$

Therefore, the volume of the gas at 127.0°C is 321.3 mL

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