Q =nRT ln (V2/V1); what is it, how it's different from W= nRT ln (V2/V1)?

The equation Q= nRT ln (V₂/V₁) represents the energy change of an ideal gas during an isothermal process.

It is an important equation to understand the behavior of ideal gases in thermodynamic systems.

Therefore, we have tried to explain everything related to Q= nRT ln (V₂/V₁) in this article, along with plenty of examples on how to use it.

So, what are you waiting for? Dive into the article and continue learning! 

What are the components of Q =nRT ln (V₂/V₁)?

In Q = nRT ln (V₂/V₁):

• Q = The change in heat or thermal energy of the gas (Units: J)
• n = Number of moles of gas (Units: mol)
• R= Ideal gas constant. The value of R stays fixed at R= 8.314 J/K.mol
• T= Absolute temperature (Units: K)
• V₂ = Final volume occupied by the gas (Units: L or m³)
• V₁ = Initial volume occupied by the gas (Units: L or m³)

An isothermal process is one in which the temperature of the system stays constant.

As per Q = nRT ln (V₂/V₁), at a fixed temperature (T), the volume of a specific number of moles of gas (n) changes from V₁ to V₂ when a certain amount of heat (Q) enters or leaves the system.

We already know from our previous knowledge that the expansion of a gas occurs (V₂ > V₁) when heat enters the system (+Q). Contrarily, the contraction of a gas (V₂ < V₁) releases energy from the system in the form of heat (-Q).

Before learning how to use this equation, let us explore the origins of Q=nRT ln (V₂/V₁). 

Derivation of Q =nRT ln (V₂/V₁)

As per the 1^st law of thermodynamics, the total energy of a system stays conserved.

dU = dQ – dW……..equation (i)

Where;

• dU = change in internal energy of a gaseous system
• Q = heat entering the system
• – dW = work done by the system

In an isothermal process, there is no change in the internal energy of the system, so dU = 0 hence,

dQ – dW = 0…..equation (ii)

Work done = dW = PdV….equation (iii)

Substituting the value of dW from equation (iii) into equation (ii) gives us:

0 = dQ – PdV……equation (iv)

The ideal gas equation is:

PV = nRT

Making P the subject of the formula:

P = nRT/V

Substituting this value of P into equation (iv):

0 = dQ – (nRT/V) dV

Making dQ the subject of the formula:

dQ = nRT/V dV ………equation (v)

Integrating both sides of equation (v) results in the final equation, i.e., Q = nRT ln (V₂/V₁)

Q = nRT (ln V₂ – ln V₁)

∴ Q = nRT ln (V₂/V₁)

Where and how to use Q =nRT ln (V₂/V₁)? – Examples

The equation Q = nRT ln (V₂/V₁) is primarily applied to find the value of heat or thermal energy of a system (Q) if all the other variables (n, T, V_1, and V₂) are given.

Furthermore, it can also be rearranged in the following different ways, depending upon which variable is unknown.

However, you must remember that consistency in units is very important. The temperature (T) must be in the absolute scale (Kelvin), while both V₁ and V₂ should be in the same units while using the equation Q = nRT ln (V₂/V₁).  

Now let us see some examples on the application of Q = nRT ln (V₂/V₁).

For example, Calculate the heat transferred (Q) during the isothermal expansion of 2 moles of a gas at 300 K, such that the volume of gas changes from 5 L to 15 L.

The information obtained from the question statement is:

• No of moles of gas = n = 2 mol
• Temperature = T = 300 K
• Initial volume = V₁ = 5 L
• Final volume = V₂ = 15 L
• Heat transferred = Q = ?

We already know that the ideal gas constant R = 8.314 J/K.mol

Therefore, substituting the above data into the equation Q = nRT ln (V₂/V₁) gives us:

⇒ Q = nRT ln (V₂/V₁)

Q = (2)(8.314)(300) ln (15/5)

Q = 4988.4 ln (3)

Find the value of ln 3 using your scientific calculator:

Q = 4988.4 (1.0986)

Q = 5480.3 J

Converting the answer from Joules to Kilojoules gives us:

1 J = 10^-3 kJ

∴ Q = 5480.3 x 10^-3 = 5.48 kJ

Result: The heat transferred during the isothermal expansion of the gas in this example is 5.48 kJ.

Another example is the – 4.3 kJ energy transfer occurred during the isothermal expansion of 1 mole of an ideal gas at T= 500 K. Use the equation Q = nRT ln (V₂/V₁) to find the value of the final volume of the gas (V₂) if its initial volume (V₁) is 100 L.

As per the question statement:

• Heat transferred = Q = 4.3 kJ = 4300 J (Converting kJ to J to bring consistency in units)
• No of moles of gas = n = 1 mol
• Temperature = T = 500 K
• Initial volume = V₁ = 100 L
• Final volume = V₂ = ?

We already know that the ideal gas constant R = 8.314 J/K.mol

Substituting the above data into the equation Q = nRT ln (V₂/V₁) gives us:

4300 = (1) (8.314) (500) ln (V₂/100)

4300 = 4157 ln (V₂/100)

4300/4157 = ln (V₂/100)

1.034 = ln (V₂/100)

As ln x = 2.303 log x, so the above equation can be rewritten as:

1.034 = 2.303 log (V₂/100)

1.034/2.303 = log (V₂/100)

0.449 = log (V₂/100)

Taking antilog on the above equation:

10^0.449= V₂/100

2.81 = V₂/100

∴V₂ = 281 L

Result: The final volume of gas in this example is 281 L.

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