Question

If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What is the volume?

Answer 1

Answer:
`1.54m^3`

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 at STP = 1 atm

`P_2` = final pressure of gas = 2.67 atm

`V_1` = initial volume of gas =
`3.61m^3`

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas at STP =
`0^oC=273+0=273K`

`T_2` = final temperature of gas =
`37.9^oC=273+37.9=310.9K`

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

`(1atm* 3.61m^3)/(273K)=(2.67* V_2)/(310.9K)`

`V_2=1.54m^3`

Thus the final volume will be
`1.54m^3`