Thermodynamic Problem: Work Done at Constant Atmospheric Pressure

[shaiqbashir]

Hi Guys!

please help me in deriving this:

A container of volume $$V_{c}$$ contains gas at a high temperature. The container id connected through a stopcock and a capillary tube to a cylinder equipped with a frictionless piston. When the stopcock is opened, the gas enters the cylinder while at the same time the pressure in the cylinder remains constant at the atmospheric value $$P_{a}$$.

a) show that the amount of work done at the end of the process is

$$W=P_{a}(V_{a}-V_{c})$$

where $$V_{a}$$ is the volume of the gas at the atmospheric temperature and pressure.

b) How much work would have been done if the gas leaked directly into the atmosphere?

Thanks in advance

 

[Navin A S]

.For part a), the amount of work done is equal to the change in the internal energy of the system. Since the pressure remains constant during the process, the work done is given by the equation W = P_{a}*ΔV, where ΔV is the change in volume. Since the volume of the container before the gas enters is V_{c}, and the volume of the gas at the atmospheric temperature and pressure is V_{a}, we can write the equation as W = P_{a}(V_{a}-V_{c}). For part b), the work done would be zero since the pressure of the system would remain constant.

 

[ravenprp]

!

Hi there! To derive the equation for the work done in this process, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, since the process is taking place at constant atmospheric pressure, the work done is equal to the pressure times the change in volume.

a) So, using the first law of thermodynamics, we can write:

ΔU = Q - W

Where ΔU is the change in internal energy, Q is the heat added, and W is the work done. We know that the process is taking place at constant atmospheric pressure, so we can rearrange the equation to solve for W:

W = Q - ΔU

Now, we need to consider the different components of this equation. Q represents the heat added, which in this case is equal to zero since the process is adiabatic (no heat is added or removed). ΔU represents the change in internal energy, which is equal to the change in enthalpy (ΔH) at constant pressure. So, we can rewrite the equation as:

W = ΔH

The change in enthalpy can be calculated using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Since the pressure is constant, we can rearrange the equation to solve for ΔH:

ΔH = nRΔT

Now, we need to consider the change in temperature (ΔT) in this process. The gas starts at a high temperature and then cools down as it expands into the cylinder. We can calculate the change in temperature using the equation ΔT = T_{a} - T_{c}, where T_{a} is the atmospheric temperature and T_{c} is the initial temperature in the container. So, we can rewrite the equation for ΔH as:

ΔH = nRT_{a} - nRT_{c}

Now, we can substitute this into our equation for work:

W = nRT_{a} - nRT_{c}

Finally, we can rearrange the equation to solve for the work done:

W = P_{a}(V_{a} - V_{c})

Where P_{a} is