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

470 °C

Step-by-step explanation:

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

`(V_(1))/(T_(1)) =(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}(V_(1))/(T_(1))& =&(V_(2))/(T_(2))\\\\ \frac{\text{20 L}}{\text{373.15 K}} &=&\frac{\text{40 L}}{T_(2)}\\\\{\text{15 000 K}} & = & 20T_(2)\\T_(2) & = &\frac{\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