Question

A gas has a volume of 1.8Lat−26◦Cand 147 kPa. At what temperature would the gas occupy 1.33 L at 217 kPa?

Answer 1

Answer: At temperature of 269 K the gas would occupy 1.33 L at 217 kPa

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 = 147 kPa

$P_2$ = final pressure of gas = 217 kPa

$V_1$ = initial volume of gas = 1.8 L

$V_2$ = final volume of gas = 1.33 L

$T_1$ = initial temperature of gas = $-26^oC=273-26=247K$

$T_2$ = final temperature of gas = ?

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

$\frac{147kPa\times 1.8L}{247K}=\frac{217kPa\times 1.33L}{T_1}$

$T_2=269K$

Thus at 269 K temperature the gas would occupy 1.33 L at 217 kPa