Question

A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bar and 100oC, and the other side, representing 2/3 of the tank, is evacuated. The membrane is ruptured and the gas fills the tank. (a) What is the final temperature of the gas? How much work is done? Is the process reversible? (b) How much work is done if the gas is returned to its original state by a reversible process? Assume nitrogen ideal gas for which Cp = (7/2) R and Cv = (5/2)R.​

Answer 1

The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.

We need to understand the thermodynamic processes in order to locate the solution.

How can I determine the gas's final temperature?

• Thermodynamic processes are any actions that result in modifications to a system's thermodynamic coordinates.

• Given that the tank is stiff and non-conducting, the answer to the question is that dQ=0.

• Without using any external force, the membrane is torn; hence, dW=0.

• The first law of thermodynamics is expressed as follows:

                    $dU=dQ-dW$     , It is 0 here.

• As we are aware,

                  $dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K$

As a result, the system's final temperature will be equal to its starting temperature.

How much work is expended?

• The process is isothermal, as we discovered.

• As a result, the work will be,

                  $W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules$

R is the gaseous universal constant.

A reversible process is what?

• Reversible processes are any operations that have the ability to be reversed.

• The system goes through the exact same states as it did during the direct procedure throughout this time.

Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.

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Answer 2

The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

How to find the final temperature of the gas?

• Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
• In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
• The membrane is raptured without applying any external force, thus, dW=0.
• We have the first law of thermodynamic expression as,

                                $dU=dQ-dW$

• Here it is zero.

                                  $dU=0$,

• As we know that,

                             $dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2$

• Thus, the final temperature of the system will be equal to the initial temperature,

                          $T_1=T_2=100^0C=373K$

How much work is done?

• We found that the process is isothermal,
• Thus, the work done will be,

                               $W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J$

Where, R is the universal gas constant.

What is a reversible process?

• Any process which can be made to proceed in the reverse direction is called reversible process.
• During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

Learn more about thermodynamic processes here:

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