Question

He gas is confined to a piston and cylinder with a mechanical stop that prevents the cylinder from expanding. The temperature of the gas is 298 K, the volume of gas in the cylinder is 1.20 L and its pressure is 2.15 bar. The ideal gas EOS is very accurate for He under these conditions.

The mechanical stop is a nut on a threaded rod that can be turned, slowly allowing the gas in the cylinder to expand. This process was carried out at a constant external pressure of 1.12 bar, maintaining thermal equilibrium at 298 K between the cylinder and the surroundings until mechanical equilibrium was reached and the He in the cylinder was also at a pressure of 1.12 bar. Further turning of the threaded nut did not result in any increase in volume.

a) Is this a reversible or irreversible process?
b) What is the work (in J)? Is the system doing work on the surroundings, or is the surroundings doing work on the system?
c) For a gas like He, which behaves ideally, internal energy depends only on the temperature of the gas. Since the temperature is constant throughout the expansion, deltaU =0. What is q (in J)?

Answer 1

  

  If a process is carried in such a way that the difference between the internal pressure and the external pressure is negligible i.e. Infinitesimal difference between the external and internal pressure then the process is called a reversible process. For expansion the requisite condition is that the external pressure must be smaller than the internal pressure , so for a reversible expansion process the external pressure must be infinitesimal smaller than the internal pressure during the course of the expansion process. But in the above said process the gas is expanding against a constant pressure of 1.12 bar where the internal pressure of the gas is 2.15 bar. So there is huge difference between the internal and the external pressure of the gas. That is why there is no reversiblity in the expansion process carried out.

So, this is a irreversible expansion process. (ans)

  Mechanical Work involved in a given process,

  

where P_ext = external constant pressure = 1.12 bar

dV = change in volume of the gas.

Now, dV = final volume - initial volume = ( V₂ - V₁ ​​​​​​)

Initial Volume if the gas (V₁) = 1.20 L

Initial pressure of the gas (P₁)= 2.15 bar

Final pressure of the gas (P₂) = 1.12 bar

Say final volume = V₂

As the gas behaves ideally and expand under constant temperature condition ,

Therefore at constant temperature from Boyle's law,

P₁V₁=P₂V₂

Or, V₂ = P₁V₁/P₂

Or, V₂ = (2.15 bar)*(1.20 L) / (1.12 bar)

Or, V₂ = 2.30 L

Therefore, dW = - P_ext dV

Or, dW = - (1.12 bar)*(2.3 L - 1.2 L)

Or, dW = - 1.23 bar L

Or, dW = - 1.23 *( 1/1.01325) atm L  

  [ As, 1 atm = 1.01325 bar ]

Or, dW = - 1.214 L atm

Now, 8.314 J = 0.082 L atm

Therefore, 1 L atm = ( 8.314 / 0.082 ) J

Or, 1 L atm = 101.39 J

Therefore, dW = - 1.214 L atm

Or, dW = - 1.214*101.39 J = - 123.08 J

Or, dW - 123.1 J   (ans)

Here the gas expand against the external constant pressure. So work is done by the system on the surrounding. It is also proved from the negative sign of the work involve.

As the gas behaves ideally, its internal energy is only depends on the temperature of the gas. But the gas is in a thermal equilibrium throughout the process i.e. its temperature remains constant throughout the expansion process. So its internal energy (U) also remains constant throughout the expansion process. So the change in internal energy (dU) is zero i.e. dU = 0

From 1st law of thermodynamics,

dU = dq +dw

Or, 0 = dq + ( - 123.1 J)

Or, dq = 123.1 J (ans)