The energy balance for one mole of gas in the system is $$ [1] \quad \Delta U = Q + W $$ where \( \Delta U \) is the change to internal energy (kJ/mol), \( Q \) is heat added to the system (kJ/mol), and \( W \) is work performed on the system (kJ/mol). Work is $$ [2] \quad W = - P \Delta V $$ where \( P \) is external pressure (Pa) and \( V \) is the specific gas volume (L/mol). Combining equations [1] and [2] yields, $$ [3] \quad \Delta U + P \Delta V = Q $$ and since \( P \) is constant, this becomes $$ [4] \quad \Delta U + \Delta ( P V ) = Q $$ and since \( H = U + PV \), where \( H \) is enthalpy (kJ/mol), $$ [5] \quad \Delta H = Q $$ For an ideal gas, $$ [6] \quad \Delta H = C_{P} \; \Delta T $$ so $$ [7] \quad Q = C_{P} \; \Delta T $$ where \( C_{P} \) is the constant-pressure heat capacity of the gas. For an ideal diatomic gas, $$ [8] \quad C_{P} = 3.5 R $$ where \( R \) is the ideal gas constant, which is 8.314x10-3 kJ/(mol K). The total gas volume (\( V_{tot} = nV \)) is calculated using the ideal gas law $$ [9] \quad PV_{tot} = nRT $$ where \( n \), the number of moles of gas, is equal to 1.0.
The energy balance for one mole in the system is $$ [1] \quad \Delta U = Q + W $$ Where \( \Delta U \) is the change to internal energy (kJ/mol), \( Q \) is heat added to the system (kJ/mol), and \( W \) is work performed on the system (kJ/mol). For a constant-volume system, \( W = 0 \) because \( \Delta V = 0 \) and $$ [2] \quad \Delta U = Q $$ For one mole of an ideal gas, \( \Delta U = C_{V} \Delta T \), where \( C_{V} \) is the constant-volume heat capacity (2.5R, where R is the ideal gas constant) [kJ/(mol K)] and heat added per mole is $$ [3] \quad Q = C_{V} \Delta T $$ The total heat added \( Q_{total} \) is $$ [4] \quad Q_{total} = n C_{V} \Delta T $$ where \( n \) is the number of moles of gas. Thus, the temperature change can be calculated from \( Q_{total} \), and the pressure calculated from the ideal gas law \( ( PV_{tot} = nRT ) \), where \( V_{tot} \) is the total volume (L) and R is 8.206x10-2 (L atm)/(mol K).
The differential energy balance for one mole of the system is $$ [1] \quad dU = dQ + dW $$ where \( dU \) is the differential change in internal energy (kJ/mol), \( dQ \) is differential heat added to the system (kJ/mol), and \( dW \) is differential work performed on the system (kJ/mol). For an adiabatic system, \( Q = 0 \), so $$ [2] \quad dU = dW $$ since \( dW = -PdV \) where \( P \) is pressure and \( V \) is the specific volume (L/mol), and for an ideal gas, \( dU = C_{V} dT \) where \( C_{V} \) is the constant-volume heat capacity (kJ/[mol K]) and equation [2] becomes $$ [3] \quad C_{V} \; dT = - P dV $$ Pressure in this equation is replaced using the ideal gas for one mole \( ( PV = RT ) \), where \( R \) is the ideal gas constant, and rearranging yields $$ [4] \quad \frac{ C_{V} }{ T } dT = - \frac{ R }{ V } dV $$ Integrating both sides between initial conditions \( ( T_{1}, V_{1} ) \) and final conditions \( ( T_{2}, V_{2} ) \) yields $$ [5] \quad C_{V} \; \mathrm{ ln } \left( \frac{ T_{ 2 } }{ T_{ 1 } } \right) = - R \; \mathrm{ ln } \left( \frac{ V_{ 2 } }{ V_{ 1 } } \right) $$ Taking exponents of both sides yields $$ [6] \quad \frac{ T_{2} }{ T_{1} } = \left( \frac{ V_{ 1 } }{ V_{ 2 } } \right)^{ R / C_{V} } $$ The ideal gas law can be used to replace \( V_1 \) and \( V_2 \) to yield $$ [7] \quad \frac{ T_{2} }{ T_{1} } = \left( \frac{ P_{ 2 } }{ P_{ 1 } } \right)^{ R / C_{P} } $$ where the constant-pressure heat capacity \( C_P = C_V + R \). For an ideal diatomic gas, $$ [8] \quad C_P = 3.5 R $$ and \( R \) is 8.314x10-3 kJ/[mol K]. In this simulation, the initial temperature \( T_{1} \) is 273 K and the initial pressure \( P_{1} \) is 1 atm. The total volume is \( V_{tot} = nV \) where \( n \) is the total number of moles.
The significance of the "external spring" scenario is to show that it is possible for pressure, temperature, and volume to all increase (or decrease) simultaneously when heat is added (or removed). The differential energy balance for the total moles of the system is $$ [1] \quad n d U = dQ_{total} + dW_{total} = n dQ - n P dV $$ where \( d U \) is the differential change in internal energy (J/mol), \( dQ_{total} \) is differential total heat added (J), \( dQ \) is differential change in total specific heat added (J/mol), \( dW_{total} \) is differential total work performed on the system (J), \( n \) is number of moles, \( P \) is pressure (N/m2), and \( dV \) is the differential change in specific volume (m3/mol). The displacement of a spring is determined with Hooke's law: $$ [2] \quad F = k \Delta z $$ where \( k \) is the spring constant (N/m), \( F \) is the force on the spring and \( \Delta z \) is the spring displacement (m). The total pressure is atmospheric pressure (\( P_{atm} \)) plus the pressure (\( P_{spr} \)) due to the force exerted by the spring on the piston: $$ [3] \quad P = P_{atm} + P_{spr} = P_{atm} + \frac{F}{A} = P_{atm} + \frac{k \Delta z}{A} $$ solving for \( \Delta z \) yields $$ [4] \quad \Delta z = \frac{ (P - P_{atm}) A}{ k } $$ The total volume change \( \Delta V_{total} \) is $$ [5] \quad \Delta V_{total} = n (V - V_{init}) = A \Delta z = \frac{ (P - P_{atm}) A^{2} }{ k } $$ where \( V_{init} \) is the initial specific volume (m3/mol). The specific volume is then: $$ [6] \quad V = \frac{ (P - P_{atm}) A^{2} }{ k \, n } + V_{init} $$ The pressure is related to the specific volume by the ideal gas law: $$ [7] \quad P = \frac{RT}{V} $$ The pressure exerted by the spring (from equations [3] and [5]) is: $$ [8] \quad P_{spr} = \frac{k \Delta z}{ A } = \frac{ k n ( V - V_{init} ) }{ A^{2} } $$ Subsituting eqn [8] and given that \( dU = C_{V} dT \) (where \( C_{V} \) is the constant volume heat capacity, 2.5R) into the energy balance (eqn [1]) and writing the equation in integral form yields: $$ [9] \quad \int_{ T_{init} }^{ T } n C_{v} dT = Q_{total} - \int_{ V_{init} }^{ V } n \left[ P_{atm} + \frac{ k \, n }{ A^{ 2 } }(V - V_{init}) \right] dV $$ where \( T_{init} \) is the initial temperature. Integrating equation [9] yields $$ [10] \quad n C_{v} (T - T_{init}) = Q_{total} - n P_{atm} (V - V_{init}) + \frac{ k n^{2} }{ A^{ 2 } } \left( \frac{ V^{2} }{ 2 } - \frac{ V_{init}^{2} }{ 2 } - V_{init} ( V - V_{init} ) \right) $$ Solving this for temperature yields $$ [11] \quad T = \frac{ Q_{total} - n \, P_{atm} (V - V_{init}) + \frac{ k n^{2} }{ A^{ 2 } } \left( \frac{ V^{2} }{ 2 } - \frac{ V_{init}^{2} }{ 2 } - V_{init} ( V - V_{init} ) \right) }{ n C_{v} } + T_{init} $$ Equations [6], [7], and [11] are solved by iteration. The calculations are carried out in SI units and the units are then converted to atm, L, and kJ in the graphic.
The differential energy balance for one mole of the system is $$ [1] \quad dU = dQ + dW $$ Where \( dU \) is the differential change to internal energy (kJ/mol), \( dQ \) is differential heat added to the system (kJ/mol), and \( dW \) is differential work performed on the system (kJ/mol). Since \( dU = C_{V} dT \) for an ideal gas, where \( C_V \) is the constant-volume heat capacity (kJ/[mol K]), and \( dT \) is zero for a constant-temperature process, then \( dU = 0 \), and the energy balance is $$ [2] \quad dQ = -dW = PdV $$ where \( P \) is the gas pressure (for a reversible process) and \( V \) is the specific gas volume (L/mol). Integrating eqn [2] yields $$ [3] \quad Q = \int_{V_{1}}^{V_{2}} P dV $$ When \( P \) is replaced using the ideal gas law for one mole \( (P = RT/V) \), then the heat added per mole is $$ [4] \quad Q = \int_{V_{1}}^{V_{2}} \frac{ RT }{ V } dV = RT \; \mathrm{ln} \left( \frac{V_{2}}{V_{1}} \right) = - RT \; \mathrm{ln} \left( \frac{P_{2}}{P_{1}} \right) $$ where \( P_2 \) is the final pressure and \( P_1 \) is the initial pressure (1 atm). The total heat added \( Q_{total} \) is $$ [5] \quad Q_{total} = - nRT \; \mathrm{ln} \left( \frac{P_{2}}{P_{1}} \right) $$ where \( n \) is the number of moles, and $$ [6] \quad V_{total} = n V $$