Question

An unknown volume of gas has a pressure of 0.50 atm and temperature of 325 K. If the pressure is raised to 1.2 atm and the temperature decreased to 320 K, giving a new volume of 48 L, what was the initial volume?

A. 65 L
B. 87 L
C. 120 L
D. 140 L

Answer 1

I have an unknown volume of a gas at a pressure of 0.5atm and a temperature of 325K. If I raise the pressure to 1.2atm, decrease the temperature to320K, and ... If 175.0mL of oxygen is collected at 23°C, what volume will the gas ... 1.2 atm x 48 L x 325 K / (0.5 atm x 320 K) = 117 L 4. You want to find the .

Answer 2

Answer : The correct option is, (C) 120 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,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 0.50 atm

`P_2` = final pressure of gas = 1.2 atm

`V_1` = initial volume of gas = ?

`V_2` = final volume of gas = 48 L

`T_1` = initial temperature of gas = 325 K

`T_2` = final temperature of gas = 320 K

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

`(0.50atm* V)/(325K)=(1.2atm* 48L)/(320K)`

`V_1=117L\approx 120L`

Therefore, the initial volume was 120 L.