An 8.0 L sample of neon gas at 300. K exerts a pressure of 900. kPa. If the gas is compressed to 2.0 L and the temperature is raised to 600. K, what will be the new pressure?

7200 kPa

Explanation:

Applying,

PV/T = P'V'/T' Equation 1

Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas

Make P' the subject of the equation

P' = PVT'/V'T Equation 2

Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K

Substitute these values into equation 2

P' = (900×8×600)/(2×300)

P' = 7200 kPa

just did the assignment and the answer was #1 +2 n