Question

The pressure in a container has decreased by 10 times, while the volume increased by 5 times from the original volume. The temperature now reads -123.15 °C. What was the starting temperature when I left (in °C)?

Answer 1

Answer: The starting temperature when I left in °C is 26.70

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = p

$P_2$ = final pressure of gas = $\frac{p}{10}$

$V_1$ = initial volume of gas = v

$V_2$ = final volume of gas = $5\times v$

$T_1$ = initial temperature of gas = ?

$T_2$ = final temperature of gas = $-123.15^oC=273+(-123.15)=149.85K$

Now put all the given values in the above equation, we get:

$\frac{p\times v}{T_1}=\frac{\frac{p}{10}\times 5\times v}{149.85K}$

$T_1=299.70K=(273-299.70)^0C=-26.70^0C$