Question

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. the gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. assume that neon behaves as an ideal gas. for the entire process, find the heat added to the gas.

Answer 1

We can use the ideal gas law PV=nRT For the first phase The starting temperature (T1) is 273.15K (0C). n is 1 mole, R is a constant, P = 1 atm, V1 is unknown. The end temperature (T2) is unknown, n= 1 mol, R is a constant, P = 3*P1= 3 atm, V2=V1 Since n, R, and V will be constant between the two conditions: P1/T1=P2/T2 or T2= (P2*T1)/(P1) so T2= (3 atm*273.15K)/(1 atm)= 3*273.15= 816.45K For the second phase: Only the temperature and volume change while n, P, and R are constant between the start and finish. So: V1/T1=V2/T2 While we don't know the initial volume, we know that V2=2*V1 and T1=816.45K So T2=(V2*T1)/V1= (2*V1*T1)/V1=2*T1= 2*816.45K= 1638.9K To find the total heat added to the gas you need to subtract the original amount of heat so 1638.9K-273.15K= 1365.75K