Question

A sample of gas is held at 100oc at a volume of 20 L.if the volume is increased to 40 L what is the new temperature of the gas in celcius.

Answer 1

Answer:

470 °C  

Explanation:

This looks like a case where we can use Charles’ Law:  

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

Data:

V₁ = 20 L; T₁ = 100 °C

V₂ = 40 L; T₂ = ?  

Calculations:

(a) Convert the temperature to kelvins

T₁ = (100 + 273.15) K = 373.15 K

(b) Calculate the new temperature

$\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{20 L}}{\text{373.15 K}} &=&\dfrac{\text{40 L}}{T_{2}}\\\\{\text{15 000 K}} & = & 20T_{2}\\T_{2} & = &\dfrac{\text{15 000 K}}{20 }\\\\T_{2} & = & \textbf{750 K}\\\end{array}$

Note: The answer can have only two significant figures because that is all you gave for the volumes.

(c) Convert the temperature to Celsius

T₂ = (750 – 273.15) °C = 470 °C