A vessel contains 2 litres of gas at a pressure of 300kPa and a temperature of 2

Question

A vessel contains 2 litres of gas at a pressure of 300kPa and a temperature of 27C .3 litres of gas at the same temperature and pressure are forced into the vessel.the temperature is then raised to 127C.Calculate the new pressure of the gas

Explanation / Answer

For all of these 2 state gas law problems do this... 1) start with P1V1 / (n1T1) = P2V2 / (n2T2) 2) identify and cancel anything constant 3) rearrange and solve for your unknown. *************** in this case, you have to sort through the problem and understand what's happening.. > you have a 2 liter container. That volume doesn't change. > but...P and T and moles change though... right? T and P change.. that's given.. and you're adding gas.. so moles changes. so... P1V1 / (n1T1) = P2V2 / (n2T2) becomes P1 / (n1T1) = P2 / (n2T2) and... P2 = P1 x (T2 / T1) x (n2 / n1) now.. you know P1 = 300 kPa T2 = 127 + 273 = 400K T1 = 27 + 273 = 300K what about n2 / n1? well... PV = nRT.. right? so.. n1 = P1V1/(RT1) and...n2 = n1 + the 3 liters of gas we added at the same P and T of the first gas... ie.. n2 = P1x2L / (RT1) + P1x3L / (RT1) = P1/(RT1) x (2L + 3L) = P1x5L / (RT1) so... n2 / n1 becomes... (P1 x 5L / (RT1)) / (P1 x 2L / (RT1)) = 5 / 2 then... P2 = P1 x (T2 / T1) x (n2 / n1) = 300kPa x (400K / 300K) x (5/2) = 1000 kPa *********** one last thing... any time you're working with volumes gases at the same temperature and pressure, you should remember that mole ratios = volume ratios... because of the PV=nRT relation @ constant T and P. A quick check of this problem... 2+3 = 5 L... 5L / 2L = 5/2..tada

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