Problem 3.102 Heating a gas at constant volume

Problem 3.102:A cylinder of fixed capacity $44.8$ liters contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15\:^oC$. Take $R$ as $8.314\:Jmol^{-1}$

Solution:

The gas is in a cylinder of fixed capacity. So the process is isochoric in nature.

The cylinder has a capacity of $44.8$ liters.

The gas is at standard temperature and pressure.

From Avogadro's law, $22.4$ liters of gas at standard temperature and pressure will be one mole.

So the given gas has $2$ mole.

Helium is monoatomic with degree of freedom $f\:=\:3$.

Specific heat capacity at constant volume, $C_v\:=\:\frac{f}{2}R$.

With $f\:=\:3$, $C_v\:=\:\frac{3}{2}R$.

The required change in temperature of the gas in the cylinder is $15\:^oC$

This is equal to change of $15\:K$.

Heat required for this constant volume process can be estimated using the relation $Q\:=\:nC_v\delta T$.

Then, $Q\:=\:2\times\frac{3}{2}R\times 15$.

Or, $Q\:=\:3\times R\times 15\:=\:3\times 8.314\times 15\:=\:374.13\:J$.

Therefore, the amount of heat needed to raise the temperature of  $44.8$ liters helium gas without changing its volume is $374.13\:J$.

3.101 Adiabatic expansion

3.101 Adiabatic expansion

Problem 3.101: A cylinder with a movable piston contains $3$ moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?

Solution:

The cylinder initially contains hydrogen at at standard temperature and pressure.

So, Initial pressure $P_1$ = $P_a$.

Initial volume $V_1$= $V$.

Initial temperature $T_1$= $T$.

Given that the cylinder is made of heat insulator and the piston is insulated by having a pile of sand on it. This prevents any exchange of heat between the hydrogen inside and the surroundings. So the process is adiabatic in nature.

When hydrogen is compressed,

Final volume $V_2$= $\frac{V}{2}$.

As the process is adiabatic, we can modify the ideal gas equation into the form $PV^{\gamma}\:=\:constant$.

The initial and final states can be related as

$P_1V_1^{\gamma}\:=\:P_2V_2^{\gamma}$

From this the final pressure, $P_2\:=\:P_1\times \frac{V_1^{\gamma}}{V_2^{\gamma}}$.

Or   $P_2\:=\:P_1\times\left(\frac{V_1}{V_2}\right)^\gamma$

Here $V_1$= $V$ and $V_2$= $\frac{V}{2}$.

Therefore, $P_2\:=\:P_1\times\left(\frac{V }{V/2}\right)^{\gamma}$

Or, $P_2\:=\:P_1\times2^{\gamma}$

Value of $\gamma$ for hydrogen is $\frac{7}{5}$.

The final pressure, $P_2\:=\:P_1\times2^{\frac{7}{5}}\:=\:2.63P_1$.

Hence the adiabatic compression will increase the pressure by an factor of $2.63$ approximately.

 

 

3.100 Work done in a cyclic process

3.100 Work done in a cyclic process

Problem 3.100: Determine work done in a cyclic process with a $P-V$ diagram as shown here.

28-03-2015 A

Solution:

Work done in acyclic process can be found by calculating the area enclosed in the $P-V$ diagram.

Here the $P-V diagram is a rectangle.28-03-2015 B

The volume varies from $V$ to $4V$ and pressure changes between $P$ and $3P$.

Longer side of the rectangle = $4V\:-\:V\:=\:3V$.

Shorter side = $3P\:-\:P\:=\:2P$.

Area of the rectangle = $3V\times 2P\:=\:6PV$.

So the work done is equal to $6PV$.

3.99 Heat exchanged in two paths

3.99 Heat exchanged in two paths

Problem 3.99: A system goes from $A$ to $B$ by two different paths in the $P-V$ diagram as shown. Heat given to the system in path $1$ is $1000\: J$. The work done by the system along path $2$ is less than path $1$ by $100 \:J$. What is the heat exchanged by the system in path $2$?

 28-03-2015 A

Solution:

For a thermodynamic process, we can apply first law in thermodynamics.

This relates the work done $W$, change in internal energy $\triangle U$ and heat energy transferred $\triangle Q$.

It is expressed as $\triangle Q\:=\: \triangle U \:+\:W$.

Or change in internal energy, $\triangle U\:=\: \triangle Q \:-\:W$.

The two processes start at the same initial state and end up in the same final state. Change in internal energy depends only on the initial and final temperature. So change in internal energy for both the path will be same.

For path $1$:

Heat given, $Q_1\:=\:1000\:J$.

Work done, = $W_1$.

Change in internal energy, $\triangle U_1\:=\: \triangle Q _1\:-\:W_1$.

Or, $\triangle U_1\:=\: 1000\:-\:W_1$......$(1)$

For path $2$:

Heat given, $Q_2$.

Work done, = $W_2\:=\:(W_1\:-\:100)$.

Change in internal energy, $\triangle U_2\:=\: \triangle Q _2\:-\:W_2$.

Or, $\triangle U_2\:=\: \triangle Q _2\:-\:(W_1\:-\:100)$. ......$(2)$

Equating change in internal energy for the two paths, we get $(2)$ = $(1)$.

Or, $\triangle U_2\:=\:\triangle U_1$.

Or, $\triangle Q _2\:-\:(W_1\:-\:100)\:=\: 1000\:-\:W_1$.

From this heat exchanged by the system in path $2$, $\triangle Q _2\:=\:(W_1\:-\:100)\:+\: 1000\:-\:W_1\:=\:900\:J$.

The heat exchanged by the system in path $2$ is equal to $900\:J$.

3.98 Change in internal energy in adiabatic process

3.98 Change in internal energy in adiabatic process

Problem 3.98: A thermally insulated vessel contains two moles of hydrogen gas. In a process its temperature changes from $200\:K$ to $300\:K$ adiabatically. Determine the change in internal energy of the gas.

Solution:27-03-2015 A

In adiabatic process, heat cannot be transferred to or from the gas. In such situation, temperature can be changed only by doing work.

If the gas is compressed adiabatically, its temperature increases.

If the volume of the container increases, gas will expand and its temperature will decrease.

From the problem it can be seen that temperature of the gas has increased.

This indicates a compression of the gas.

Hydrogen is a diatomic gas with five degrees of freedom.

Change in internal energy for any thermodynamic process is $\triangle U\:=\:nC_v\triangle T\:=\:nC_v(T_f\:-\:T_i)$.

Here $C_v$ is molar specific heat at constant volume, $n$ is the number of moles, $T_i$ is the initial temperature and $T_f$ is the final temperature.

$C_v$, the molar specific heat capacity for a diatomic gas is $\frac{5}{2}R$ where $R\:=\:8.314\:J mol^{-1}K^{-1}$ is the universal gas constant.

Now calculations:

Given that,

Number of moles = $2$.

Initial temperature = $200\:K$.

Final temperature = $300\:K$.

Therefore, change in internal energy,$\triangle U\:=\:2\times \frac{5}{2}\times8.314 (300\:-\:200)\:=\:4157\:J$.

Internal energy of the gas increases by $4157$ joule.

3.97 Adiabatic work done

3.97 Adiabatic work done

Problem 3.97: A thermally insulated vessel contains two moles of a mono-atomic gas. In a process its temperature changes from $200\:K$ to $300\:K$ adiabatically. Determine the work done in this process.

Solution:

19-03-2015 A

Heat transferred $\triangle Q$ is related to change in internal energy $\triangle U$ and work done $W$ as

$\triangle Q\:=\:\triangle U\:+\:W$.

In adiabatic process, heat is not transferred between the system and the surrounding.

Therefore, $0\:=\:\triangle U\:+\:W$.

Or $\triangle U\:=\:-W$.

Work done in adiabatic process is the negative of the change in internal energy.

Change in internal energy for any thermodynamic process is $\triangle U\:=\:nC_v\triangle T\:=\:nC_v(T_f\:-\:T_i)$.

$C_V$, the molar specific heat capacity for a mono atomic gas is $\frac{3}{2}R$ where $R\:=\:8.314\:Jmol^{-1}K^{-1}$ is the universal gas constant.

Change in internal energy,$\triangle U\:=\:nC_v(T_f\:-\:T_i)$.

Work done, $W\:=\:- \triangle U\:=\:nC_v(T_i\:-\:T_f)$.

Therefore Work done, $W\:=\:2\times \frac{3}{2}\times 8.314(200\:-\:300)\:=\:-\:2494.2\:J$.

Negative work indicates a compression of the gas.