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

Step-by-step explanation:

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

The combined gas equation is,

`(P_1V_1)/(T_1)=(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:

`(0.98* 0.165)/(239)=(1* V_2)/(273)`

`V_2=0.185l`

Thus the volume of this gas at STP be 0.185 L