Question

If some methane gas at 115 kPa and 33C is allowed to expand to 15.0dm^3 from 8.00 dm^3 when the temperature changed to 51.2 C. What pressure will the same amount of gas exert?

Answer 1

228.4 kPa

Step-by-step explanation:

From thw question,

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

Where P = Inital pressure, T = Initial Temperature, V = Initial Volume, P' = fInal Pressure, V' = Final volume, T' = Final Temperature.

Make P' the subject of the equation

P' = PVT'/TV'...................... Equation 2

Given: P = 115 kPa, V = 15.0 dm³, T = 33°C = (33+273) K = 306 K, T' = 51.2°C = 51.2+273 = 324.2 K, V' = 8.00 dm³

Substitute these values into equation 2

P' = (115×15×324.2)/(306×8)

P' = 228.4 kPa.