Question

A sample of a gas occupies 1.80 L at -10. °C and 450. Torr. What will be the temperature if the pressure is increased to 800. Torr and the volume is decreased by 1/3?

Answer 1

Answer: 156 K

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 = 450 torr

$P_2$ = final pressure of gas = 800 torr

$V_1$ = initial volume of gas = 1.80 L

$V_2$ = final volume of gas = $\frac{1}{3}\times 1.80L=0.60L$

$T_1$ = initial temperature of gas = $-10^oC=273-10=263K$

$T_2$ = final temperature of gas = ?

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

$\frac{450\times 1.80}{263}=\frac{800\times 0.60}{T_2}$

$T_2=156K$

Thus the final temperature will be 156 K