# Calculate the amount of heat added to the ideal monatomic gas

by Tab Caps
Tags: heat, ideal gas, monatomic, thermodynamic, thermodynamics
 P: 1 1. The problem statement, all variables and given/known data Three moles of an ideal monatomic gas expand at a constant pressure of 3.00 atm ; the volume of the gas changes from 3.40×10^-2 m^3 to 4.50×10^-2 m^3 Calculate the amount of heat added to the gas 2. Relevant equations pV = nRT dQ=nCvdT Cv = (3/2)R for a monatomic ideal gas 3. The attempt at a solution Using pV = nRT I found that the initial temperature Ti = 414 K and Tf = 548 K. Then I want to use the equation dQ = n(Cv)dT to solve for the amount of heat added in the system. The correct answer is 8360 J, but when I do: 3*(3/2)*R*(548-414) = 5010 J I think my problem is coming from not understanding the difference between dT and $\Delta$T, or dQ and $\Delta$ Q. I actually know Δ is just the change (final-initial), and the derivative is the infinitesimal rate of change, so perhaps I'm just trying the wrong formula since I don't have dT? Or is there a way to figure out dT from what I'm given?
 Quote by Tab Caps I think my problem is coming from not understanding the difference between dT and $\Delta$T, or dQ and $\Delta$ Q. I actually know Δ is just the change (final-initial), and the derivative is the infinitesimal rate of change, so perhaps I'm just trying the wrong formula since I don't have dT? Or is there a way to figure out dT from what I'm given?
$$\begin{array}{rcl} dQ & = & n C_V dT \\ \int_{Q_i}^{Q_f} dQ & = & n C_V \int_{T_i}^{T_f} dT \\ \left. Q \right|_{Q_i}^{Q_f} & = & n C_V \left[ T \right]_{T_i}^{T_f} \\ \Delta Q & = & n C_V \Delta T \end{array}$$