A rigid steel tank contains 100 moles of nitrogen gas at 320K and
1.00 bar pressure. The tank is heated until the pressure is 1.50 bar.
Assuming the gas behaves ideally, calculate Qsys. CV = 20.811
J/mol·K
To calculate Qsys, we need to use the equation for heat transfer:
Qsys = n * CV * (Tfinal - Tinitial)
Where:
Qsys = heat transferred to the system
n = number of moles of gas
CV = molar heat capacity at constant volume
Tfinal = final temperature
Tinitial = initial temperature
Given:
n = 100 moles
CV = 20.811 J/mol·K
Tinitial = 320 K
Pinitial = 1.00 bar
Pfinal = 1.50 bar
We need to convert pressure to absolute pressure in order to use it in the ideal gas law. Absolute pressure (Patm) = P + P0, where P0 is the atmospheric pressure. Assuming atmospheric pressure is 1.00 bar:
Pinitial = 1.00 + 1.00 = 2.00 bar
Pfinal = 1.50 + 1.00 = 2.50 bar
Now we can use the ideal gas law to find the final temperature:
PV = nRT
For the initial state:
Pinitial * V = n * R * Tinitial
(2.00 bar) * V = (100 mol) * (8.314 J/mol·K) * (320 K)
Solving for V:
V = (100 mol * 8.314 J/mol·K * 320 K) / (2.00 bar)
For the final state:
Pfinal * V = n * R * Tfinal
(2.50 bar) * V = (100 mol) * (8.314 J/mol·K) * Tfinal
Substituting the value of V from above:
(2.50 bar) * [(100 mol * 8.314 J/mol·K * 320 K) / (2.00 bar)] = (100 mol) * (8.314 J/mol·K) * Tfinal
Simplifying:
Tfinal = [(2.50 bar) * (100 mol * 8.314 J/mol·K * 320 K)] / [(2.00 bar) * (100 mol * 8.314 J/mol·K)]
Now we can calculate Qsys using the equation:
Qsys = n * CV * (Tfinal - Tinitial)
= (100 mol) * (20.811 J/mol·K) * (Tfinal - 320 K)
By plugging in the value of Tfinal and Tinitial, we can solve for Qsys.