A 0.25 mol sample of neon gas at 18 °C and 0.799 atm is heated to 49 °C. The new gas pressure is 1.11
atm. What is the change in volume of the neon gas?

The change in volume is 1.52 L

Explanation:

According to ideal gas equation:

P = pressure of gas = 0.96 atm

V = Volume of gas = ?

n = number of moles

R = gas constant =

T =temperature =

The combined gas equation is,

where,

= initial pressure of gas = 0.799 atm

= final pressure of gas = 1.11 atm

= initial volume of gas = 7.47 L

= final volume of gas = ?  

= initial temperature of gas =

= final temperature of gas =

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

The change in volume = (7.47 - 5.95)L=1.52 L