Question

A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What is the new pressure?

a. 0.374 atm
b. 0.534 atm
c. 2.14 atm
d. 1.87 atm
e. 0.468 atm

Answer 1

Answer : The new pressure is, 0.534 atm

Solution :

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 = 1.00 atm

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = 250 ml

`V_2` = final volume of gas = 500 ml

`T_1` = initial temperature of gas =
`20^oC=273+20=293K`

`T_2` = final temperature of gas =
`40^oC=273+40=313K`

Now put all the given values in the above equation, we get the final pressure of gas.

`(1atm* 250ml)/(293K)=(P_2* 500ml)/(313K)`

`P_2=0.534atm`

Therefore, the new pressure is, 0.534 atm