Question

A sample of argon has a volume of 0.165L at -34.0°C and a 0.98atm of pressure. What would the volume of this gas at STP be?

Answer 1

Answer: 0.185 L

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 = 0.98 atm

$P_2$ = final pressure of gas = 1 atm  (at STP)

$V_1$ = initial volume of gas = 0.165 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $-34.0^oC=273-34.0=239.0K$

$T_2$ = final temperature of gas = $273K$  (at STP)

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

$\frac{0.98\times 0.165}{239}=\frac{1\times V_2}{273}$

$V_2=0.185l$

Thus the volume of this gas at STP be 0.185 L