A sealed cylinder of gas contains nitrogen gas at 1.00 103 kPa pressure and a temperature of 27°C.?

The cylinder is left in the sun, and the temperature of the gas increases to 56°C. What is the new pressure in the cylinder?





the answer is expressed in kPa

A sealed cylinder of gas contains nitrogen gas at 1.00 103 kPa pressure and a temperature of 27°C.?
OK.





This is an applied ideal gas law problem, where volume and molar amount of gas is kept constant, but pressure and temperature vary.





PV=nRT Since V, n, R are constant, P= KT or P/T=K


{This is sometimes called Gay-Lussac's Law.}





Remember that temperature must be in degrees above absolute zero; for Celsius scale those degrees are in Kelvin, and are equal to 273.15 + C.


So:


27 C = 300.15 K


56 C = 329.15 K





Thus:





P/T= 103 kPa/300.15 = P/329.15





Solve for P = 112.9 kPa
Reply:PV=nRT





At the initial state, P1V=nRT1


At final state, P2V=nRT2





rearrange the equations and set it equal to each other, you will get P2=(P1/T1)*T2