First law of thermodynamics | Notes | Short answer questions discussion | Numerical problem solutions | Grade 12 NEB (Physics) | Physics in Depth

First law of thermodynamics

• Thermodynamic system is any macroscopic system (i.e., the system involving extremely large number of particles ). Thermodynamic systems may be closed (exchange of energy(heat or work) between system and the surroundings but not the matter) , open (exchange of energy and matter between the system and the surroundings) and isolated (no exchange of energy and matter between the system and the surroundings).



• Thermodynamic parameters are measurable quantities associated with the system, such as the pressure P , the volume V and the temperature T . A thermodynamic state is the set of thermodynamic parameters that are required for the description of the system. For e.g., ( P , V , T ) is the thermodynamic state that describes the thermodynamic system having pressure P , the volume V and the temperature T .



• A thermodynamic system is said to be in thermodynamic equilibrium when the thermodynamic state does not change with time.



• For a system with thermodynamic parameters P , V and, T , the work done dW by a system (say gas) in an infinitesimal transformation in which volume increases by dV is given by , \[dW=PdV\]



• Consider a thermodynamic process in which an amount dQ of heat is given to the gas and an amount dW of work is done by it. The total energy of the gas must increase by dQ-dW . If this energy does not appear as a systematic motion (i.e., the motion of a container ), then this net energy dQ-dW must go in the form of its internal energy. If dU be the change in internal energy, we get, \[dU=dQ-dW\] or, \[dQ=dU+dW ... (i)\] Equation (i) is the statement of the first law of thermodynamics . This can also be rewritten as, \[dQ=dU+PdV\]. First law of thermodynamics is also known as the principle of conservation of energy (i.e., energy can neither be created nor be destroyed, but can be changed from one form to another).



• Consider the volume of gas of mass m kept constant and heat dQ is given to it. If its temperature rises by dT , the specific heat capacity of gas at constant volume is, \[c_v=\left(\frac{dQ}{mdT}\right)_{const. volume} ... (i)\] Similarly, the specific heat capacity at constant pressure is, \[c_p=\left(\frac{dQ}{mdT}\right)_{const. pressure}... (ii)\]
  The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume is, \[C_v=\left(\frac{dQ}{ndT}\right)_{const. volume}... (iii)\] and the molar heat capacity at constant pressure is, \[C_p=\left(\frac{dQ}{ndT}\right)_{const. pressure}... (iv)\] Here, n is the number of moles of a gas.
  Since, \(n=\frac{m}{M}\) where M is the molar mass of the gas.
  From equation (iii), \[C_V=\frac{dQ}{\frac{m}{M}dT}\] Therefore, \[C_v=Mc_v\] Similarly, \[C_p=Mc_p\] The unit of specific heat capacity is J /kg-K whereas that of molar heat capacity is J/mol-K.



• Mayer's formula : \[C_p-C_v=R\] If M is the molar mass of a gas, then, \[\begin{align*} \frac{C_p}{M}-\frac{C_v}{M}&=\frac{R}{M}\\ \therefore c_p-c_v&=r \end{align*}\] Here, r is the gas constant per unit mass.



• Thermodynamic processes is said to take place when the thermodynamic parameters of the system change with time. They are:
  
  • Isothermal process
  • Adiabatic process
  • Isochoric process
  • Isobaric process
  • Reversible process
  • Irreversible process



• The process in which the pressure and the volume of a system change without any change in its temperature is called isothermal process . The equation of an isothermal process for an ideal gas is, \[PV=\text{constant}\] Since the internal energy depends only upon the temperature, dU = 0 in isothermal process.
  
  From the first law of thermodynamics, \[\begin{align*} dQ&=dU+dW\\ dQ&=dW \hspace{0.1cm} \because dU=0\\ dQ&=PdV \end{align*}\] So, the heat added to the system is equal to the work done by the system in isothermal process.
  
  Work done during an isothermal process is \[W=nRT\hspace{0.1cm} ln\left(\frac{V_2}{V_1}\right)\] Equivalently, \[W=nRT\hspace{0.1cm} ln\left(\frac{P_1}{P_2}\right)\] As dT = 0 during isothermal process, the molar heat capacity is, \[C_{isothermal}=\frac{dQ}{ndT}=\infty\]



• The process in which pressure, volume and temperature of the system change but there is no exchange of heat between the system and the surrounding is called adiabatic process . The equation of an adiabatic process for an ideal gas is, \[PV^\gamma=\text{constant}\] Here, \(\gamma\) is the ratio of molar heat capacities of a gas i.e., \(\gamma=\frac{C_p}{C_v}\).
  
  Value of \(\gamma\) :
  
  • \(\gamma\) = 1.67 for monoatomic ideal gas.
  • \(\gamma\) = 1.40 for diatomic ideal gas.
  • \(\gamma\) = 1.33 for triatomic ideal gas.
  
  Equation for an adiabatic processes can also be written as \[TV^{\gamma-1}=\text{constant}\] and, \[P^{1-\gamma}T^\gamma = \text{constant}\]
  
  From the first law of thermodynamics \[\begin{align*} dQ&=dU+dW\\ dU&=-dW \hspace{0.1cm}\because dQ=0 \end{align*}\] So, the work is done by the gas at the expense of internal energy during adiabatic process.
  
  Work done during an adiabatic process is \[W=\frac{1}{\gamma-1}(P_1V_1-P_2V_2)\]
  As dQ = 0 during adiabatic process, the molar heat capacity is then \[C_{adiabatic}=\frac{dQ}{ndT}=0\]



• The process in which the volume remains constant and no work is done by the system is called the isochoric process . \[\begin{align*} dQ&=dU+dW\\ dQ&=dU+PdV\\ dQ&=dU \hspace{0.1cm} \because dV=0 \end{align*}\]



• The process in which the pressure remains constant is called the isobaric process .



• A reversible process is one which can be retraced in the reverse order, so that it passes exactly through the same state as in the direct process. For e.g., an infinitesimally slow expansion and compression of an ideal gas at a constant pressure.



• An irreversible process is one which cannot be retraced in the opposite order by reversing the controlling factors. For e.g., rusting of iron.

Short Answer Questions Solution

1. Can we define specific heat capacity for an adiabatic process? Explain.
   Solution :
   The specific heat capacity is defined as, \[c=\frac{dQ}{mdT}\] Here, \(dQ\) is the amount of heat required to uniformly raise the temperature of the body by a small increment \textit{dT}.
   Since, in adiabatic process, there is no exchange of heat between the system and the surrounding (i.e., \textit{dQ=0}), the specific heat capacity is found to be zero.
   Thus, specific heat capacity for an adiabatic process is zero but defined.



2. Explain why C\(_p\) is always greater than C\(_v\)?
   Solution :
   The molar heat capacity of a gas at constant volume is, \[C_v=\left(\frac{dQ}{n dT}\right)_v\] and the molar heat capacity of a gas at constant pressure is, \[C_p=\left(\frac{dQ}{ndT}\right)_p\] In case of \(C_v\), the volume is constant and all heat energy supplied to the system is used to increase the internal energy only and so the temperature of the gas. But in case of \(C_p\), the pressure is constant, and volume changes. So, the heat energy is required not only to increase the internal energy but also to do work against the external pressure. So, more heat will be required for increasing the temperature of the gas when the gas is heated through the same temperature. Hence, \(C_p\) is greater than \(C_v\).



3. State first law of thermodynamics and write it's two limitations.
   Solution :
   First law of thermodynamics states that, energy can neither be created nor be destroyed but it changes from one form to another i.e., energy of an isolated system remains constant.
   Mathematically, \[dQ=dU+dW\] where, dQ is the quantity of heat supplied to a system, dW is the amount of external work done and dU is the increase in internal energy of the molecules.
   The limitations of first law of thermodynamics are:
   
   • It does not indicate the direction of heat transfer.
   • It does not give to what extent the mechanical energy is obtained from the heat energy.



4. When a gas expands adiabatically, it does work on its surroundings. But, if there is no heat input to the gas where does the energy come from?
   Solution :
   From first law of thermodynamics, \[dQ=dU+dW\] In adiabatic process, \(dQ=0\), so, \[-dW=dU\]\ Hence, the gas does work on its surrounding at the expense of its internal energy during the adiabatic process.



5. When you blow air out of your mouth wide open, it feels hot but when you purse your lips and breathe air out, it feels cold. Why?
   Solution :
   When we blow air out with mouth wide open, the air coming from mouth is at, or close to, body temperature, so it feels warm. However, when we blow air by pursing our lips, the adiabatic expansion of air takes place. From the relation, \[PV^\gamma = \text{constant}\] We find from above relation that due to the decrease in pressure outside the mouth, volume of air expands . Similarly from the relation, \[TV^{\gamma-1}=\text{constant}\] We find that, as the volume increases, the temperature of the air falls down and the air cools. Hence, the air feels cold when we blow air by pursing our lips due to the adiabatic exxpansion of air.



6. Define reversible and irreversible process with examples.
   Solution :
   Reversible process:
   A reversible process is one which can be retraced in the opposite order, so that it passes exactly through the same state as in the direct process. If work is done by the substance in the direct process, an equal amount of work will be done on the substance in the reverse process. Examples are: infinitesimally slow expansion and compression of ideal gas at constant pressure, all thermal processes taking place at infinitesimally slow rate, etc.
   Irreversible process:
   An irreversible process is one which cannot be retraced in the opposite order by reversing the controlling factors. Examples are: Rusting of iron, decay of matter, etc.

Numerical Problem Solutions

1. Gas in a cylinder, initially at a temperature of 17\(^0\) C and a pressure of 1.01\(\times\)10\(^5\) Nm\(^{-2}\) is to be compressed to one eight of its volume. What would be the difference between the final pressures if the compression were done (a) isothermally (b) adiabatically. What would be the final temperature in the latter case? (Ratio of the molar heat capacities=1.40)
   Solution :
   initial temperature, \(T_1\) = 17 + 273 = 290 K
   initial pressure, \(P_1\) = 1.01\(\times\)10\(^5\) Nm\(^{-2}\)
   ratio of molar heat capacities, \(\gamma\) = 1.40
   initial volume , \(V_1\) be V.
   final volume, \(V_2\) = \(\frac{V}{8}\)
   
   When the gas is compressed isothermally,
   Let, \(P_2\) be the final pressure and \(T_2=T_1\) be the final temperature of the gas.
   When the gas is compressed adiabatically,
   Let, \(P_2'\) be the final pressure and \(T_2'\) be the final temperature of the gas.
   We have to find the difference between the final pressures and also find the final temperature in the case of adiabatic compression.
   
   (a) For isothermal compression,
   
   The equation of an isothermal process is
   \[\begin{align*} P_1V_1&=P_2V_2\\ P_2&=\frac{P_1V_1}{V_2}\\ &=\frac{1.01 \times 10^5 \times V}{\frac{V}{8}}\\ &=1.01 \times 10^5 \times 8\\ \therefore P_2&=8.08 \times 10^5\hspace{0.1cm}\text{Nm}^{-2} \end{align*}\] (b) For adiabatic compression,
   
   The equation of an adiabatic process is
   \[\begin{align*} P_1 V_1^{\gamma}&=P_2' V_2^{\gamma}\\ P_2'&=\frac{P_1 V_1^{\gamma}}{V_2^{\gamma}}\\ &=1.01 \times 10^5 \times \left(\frac{V}{\frac{V}{8}}\right)^{\gamma}\\ &=1.01 \times 10^5 \times (8)^{1.4}\\ \therefore P_2'&=1.85\times 10^6\hspace{0.1cm}\text{Nm}^{-2} \end{align*}\] So, the difference between the final pressures is \[P_2'-P_2=1.04 \times 10^6\hspace{0.1cm}\text{N/m}^{-2}\] Now, the final temperature for the adiabatic case is calculated as, \[\begin{align*} T_1 V_1^{\gamma-1}&= T_2' V_2^{\gamma-1}\\ T_2'&=T_1 \times \left(\frac{V}{\frac{V}{8}}\right)^{\gamma-1}\\ T_2'&=290 \times (8)^{1.4-1}\\ \therefore T_2'&=666.24 \hspace{0.1cm}\text{K} \end{align*}\]



2. A mass of air occupying initially a volume of 2\(\times\)10\(^{-3}\) m\(^3\) at a pressure of 760 mm of Hg and a temperature of 20.0 \(^0\) C is expanded adiabatically and reversibly to twice its volume, and then compressed isothermally and reversibly to a volume of 3\(\times\)10\(^{-3}\) m\(^3\). Find the final temperature and pressure, assuming the ratio of the specific heat capacities of air to be 1.40.
   Solution :
   initial volume, \(V_1\) = 2\(\times\)10\(^{-3}\) m\(^3\)
   initial pressure, \(P_1\) = 760 mm of Hg = 1.01 \(\times \) 10\(^5\) Nm\(^{-2}\)
   initial temperature, \(T_1\) = 20.0 \(^0\) C = 273 + 20 K = 293 K
   
   When the air is expanded adiabatically,
   Let, \(V_2\) be the final volume, \(P_2\) be the final pressure and \(T_2\) be the final temperature.
   So, final volume, \(V_2\) = 2 \(V_1\) = 4 \(\times\) 10\(^{-3}\) m\(^3\)
   When the air is compressed isothermally,
   Let, \(V_2'\) be the final volume, \(P_2'\) be the final pressure and \(T_2'=T_2\) be the final temperature.
   So, final volume, \(V_2'\) = 3 \(\times\) 10\(^{-3}\) m\(^3\)
   
   \(\bullet\)For adiabatic expansion,
   
   \[\begin{align*} P_1 V_1^\gamma & = P_2 V_2 ^\gamma\\ P_2& =P_1 \left(\frac{V_1}{V_2}\right)^\gamma\\ &=1.01 \times 10^5 \times \left(\frac{2\times 10^{-3}}{4 \times 10^{-3}}\right)^{1.4}\\ &=1.01 \times 10^5 \times (0.5)^{1.4}\\ \therefore P_2&=38271.84 \hspace{0.1cm} \text{Nm}^{-2} \end{align*}\] Now,
   \[\begin{align*} T_1V_1^{\gamma-1}&=T_2 V_2 ^{\gamma-1}\\ T_2&=T_1 \times \left(\frac{V_1}{V_2}\right)^{\gamma-1}\\ &=293 \times \left(\frac{2 \times 10^{-3}}{4 \times 10^{-3}}\right)^{1.4-1}\\ &= 293 \times (0.5)^{0.4}\\ \therefore T_2 &=222.05 \hspace{0.1cm} \text{K} \end{align*}\]
   \(\bullet\) For isothermal compression,
   \[\begin{align*} P_2 V_2 & = P_2' V_2'\\ P_2'&=\frac{P_2 V_2}{V_2'}\\ &=\frac{38271.84\times 4\times 10^{-3}}{3 \times 10^{-3}}\\ \therefore P_2'&=51029. 12 \hspace{0.1cm} \text{Nm}^{-2} \end{align*}\] And the final temperature in the case of isothermal compression is equal to the initial temperature as during isothermal process, there is no change in temperature.



3. Air initially at 27\(^0\) C and 750 mm of Hg pressure is compressed isothermally until its volume is halved. It is then expanded adiabatically until its original volume is recorded. Assuming the change to be reversible, find the final temperature and pressure.
   Solution :
   initial temperature, \(T_1\) = 27\(^0\) C = 273 + 27 = 300 K
   initial pressure, \(P_1\) = 750 mm of Hg = 0.750 \(\times\) 13600 \(\times\) 9.8 = 99960 Nm\(^{-2}\) \(\because P=\rho g h\)
   
   Let, the initial volume (i.e., \(V_1\) ) be V.
   When the gas is compressed adiabatically, its temperature remains constant and the volume in our case has halved.
   \[V_2=V\] Let \(P_2\) be the final pressure during isothermal process and the final temperature \(T_2=T_1=300 \hspace{0.1cm}\text{K}\)
   When the gas is expanded adiabatically,
   \[V_2'=V\] Let \(P_2'\) be the final pressure during adiabatic process and \(T_2'\) be the final temperature.
   \(\bullet\) During isothermal compression,
   \[\begin{align*} P_1 V_1 &= P_2 V_2\\ P_2 &=P_1 \times \frac{V_1}{V_2}\\ &=99960 \times \frac{V}{\frac{V}{2}}\\ &=99960 \times 2\\ \therefore P_2 &=199920 \hspace{0.1cm}\text{Nm}^{-2} \end{align*}\] Now,
   \(\bullet\) During adiabatic expansion,
   \[\begin{align*} P_2'V_2'^\gamma&=P_2 V_2^\gamma\\ P_2'&=P_2 \times \left(\frac{V_2}{V_2'}\right)^\gamma\\ &=199920 \times \left(\frac{\frac{V}{2}}{V}\right)^{1.4}\\ &=199920 \times (0.5)^{1.4}\\ P_2'&=75755.51\hspace{0.1cm}\text{Nm}^{-2} \end{align*}\] Also, \[\begin{align*} T_2' V_2'^{\gamma-1}&= T_2 V_2^{\gamma-1}\\ T_2'&=T_2 \left(\frac{V_2}{V_2'}\right)^{0.4}\\ &=300 \times (0.5)^{0.4}\\ \therefore T_2'&=227.4 \hspace{0.1cm}\text{K} \end{align*}\]



4. How much work is done by an ideal gas in an isothermal compression of 30 litres at 1.01\(\times\)10\(^5\) Nm\(^{-2}\) to 3 litres?
   Solution :
   initial volume, \(V_1\) = 30 l = 30 \(\times \) 10 \(^{-3}\) m\(^3\)
   final volume, \(V_2\) = 3 l = 3\(\times \) 10 \(^{-3}\) m\(^3\)
   pressure, P = 1.01\(\times\)10\(^5\) Nm\(^{-2}\)
   Work done, W = ?
   
   Now, \[\begin{align*} W&=nRT ln \left(\frac{V_2}{V_1}\right)\\ &=PV_1 ln \left(\frac{3}{30}\right)\hspace{0.1cm}\because PV=nRT\\ &= 1.01 \times 10^5 \times 30 \times 10^{-3} \times (-2.30)\\ \therefore W&=-6969 \hspace{0.1cm}\text{J} \end{align*} \]



5. A sample of gas (\(\gamma\)=1.5) is taken through an adiabatic process in which the volume is compressed from 1600 cm\(^3\) to 400 cm\(^3\). If the initial pressure is 150 KPa, a) what is the final pressure and b) how much work is done by the gas in the process?
   Solution :
   initial volume, \(V_1\) = 1600 cm\(^3\) = 16 \(\times\) 10\(^{-4}\) m\(^3\)
   final volume, \(V_2\) = 400 cm\(^3\) = 4 \(\times\) 10\(^{-4}\) m\(^3\)
   initial pressure, \(P_1\) = 150 KPa = 150 \(\times \) 1000 = 1.5 \(\times \) 10\(^5\) Pa
   (a) final pressure, \(P_2\) = ?
   (b) work done by the gas, W = ?
   
   For an adiabatic process,
   
   \[\begin{align*} P_1V_1^\gamma&=P_2V_2^\gamma\\ P_2&=P_1\left(\frac{V_1}{V_2}\right)^\gamma\\ &=1.5 \times 10^5 \times \left(\frac{16}{4}\right)^1.5\\ &=1.2 \times 10^6 \hspace{0.1cm}\text{Pa} \end{align*}\]
   Now, the work done is
   \[\begin{align*} W&=\frac{1}{\gamma-1}\times (P_1V_1-P_2V_2)\\ &=\frac{1}{0.5}\times (1.5 \times 10^5 \times 16 \times 10^{-4}-1.2 \times 10^6 \times 4 \times 10^{-4})\\ \therefore W &=-480 \hspace{0.1cm}\text{J} \end{align*}\]



6. A certain volume of dry air at NTP is allowed to expand five times of its original volume under adiabatic condition. Calculate the final pressure and temperature.
   Solution :
   There are very contradictory values of standards in normal temperature and pressure. According to NIST (National institute of science and technology, U.S.), at NTP, temperature is 20\(^\circ\) C and pressure is 1.01 \(\times\) 10\(^5\) Pa.
   
   initial temperature, \(T_1\) = 20\(^\circ\) C = 293 K
   initial pressure, \(P_1\) = 1.01 \(\times\) 10\(^5\) Pa
   ratioa of molar specific heat capacities, \(\gamma\) = 1.4
   
   Let the original volume of gas, i.e., \(V_1\) be V. Then,
   
   final volume, \(V_2\) = 5 V
   final pressure, \(P_2\) = ?
   final temperature, \(T_2\) = ?
   
   \underline{For adiabatic expansion},
   
   \[\begin{align*} P_1 V_1^\gamma & = P_2 V_2^\gamma \\ P_2&=P_1 \times \left(\frac{V_1}{V_2}\right)^\gamma\\ &=1.0.1 \times 10^5 \times \left(\frac{V}{5V}\right)^{1.4}\\ \therefore P_2&=1.06 \times 10^4 \hspace{0.1cm} \text{Pa} \end{align*}\] Then,
   \[\begin{align*} T_1V_1^{\gamma-1}&=T_2V_2^{\gamma-1}\\ T_2&=T_1 \times \left(\frac{V_1}{V_2}\right)^{\gamma-1}\\ &=293 \times \left(\frac{V}{5V}\right)^{0.4}\\ \therefore T_2&=153.91 \hspace{0.1cm}\text{K} \end{align*}\]



7. For hydrogen the molar heat capacities at constant volume and constant pressure are 20.5 J/mol/K and 28.8 J/mol/K. Calculate (i) the heat needed to raise the temperature of 8 g of hydrogen from 10\(^\circ\) C to 15 \(^\circ\) C at constant pressure, (ii) the increase in internal energy of the gas [molar mass of hydrogen = 2 g]
   Solution :
   
   molar heat capacities at constant volume, \(C_v\) = 20.5 J mol\(^{-1}\) K \(^{-1}\)
   molar heat capacities at constant pressure, \(C_p\) = 28.8 J mol\(^{-1}\) K \(^{-1}\)
   (i) dQ = ? for dT = 15 - 10 = 5 \(^\circ\) C
   (ii) dU = ?
   molar mass of hydrogen, M = 2 g
   mass of hydrogen, m = 8 g
   
   (i) \[\begin{align*} dQ&=nC_p dT\\ &=\frac{m}{M} C_p dT\\ &=\frac{8}{2}\times 28.8 \times 5\\ \therefore dQ&=576 \hspace{0.1cm}\text{J} \end{align*}\] (ii) \[\begin{align*} dU&=nC_v dT\\ &=\frac{m}{M} C_v dT\\ &=\frac{8}{2}\times 20.5 \times 5\\ \therefore dU&=410 \hspace{0.1cm}\text{J} \end{align*}\]



8. If the ratio of specific heat capacities of a gas is 1.4 and its density at STP is 0.09 kgm\(^{-3}\). Calculate the values of specific heat capacities at constant pressure and at constant volume.
   Solution :
   At STP,
   
   temperature, T = 273 K
   pressure, P = 1.0 \(\times\) 10\(^5\) P
   
   Here,
   \(\gamma\) = 1.4
   density , \(\rho\) = 0.09 kgm\(^{-3}\)
   \(c_p\) = ?
   \(c_v\) = ?
   
   From an ideal gas equation,
   \[\begin{align*} PV&=nRT\\ PV&=\frac{m}{M}RT\\ P&=\frac{m}{V}\frac{R}{M} T\\ P&=\rho r T\\ 10^5&= 0.09 \times r \times 273\\ r&=\frac{10^5}{0.09\times 273}\\ r&=4070 \hspace{0.1cm}\text{J/kg/K} \end{align*}\] Also,
   \[\because \gamma=\frac{c_p}{c_v}=1.40\] \[c_p=1.40 c_v\] Then,
   \[\begin{align*} c_p-c_v&=r\\ 1.40 c_v-c_v&=4070\\ 0.40 c_v&=4070\\ c_v&=\frac{4070}{0.40}\\ \therefore c_v&=1.01 \times 10^4 \hspace{0.1cm}\text{J/kg/K} \end{align*}\] And, \[c_p=1.40 c_v=1.40 \times 1.01 \times 10^4=1.41\times 10^4 \hspace{0.1cm}\text{J/kg/K}\]



9. The density of an ideal gas is 1.6 kgm\(^{-3}\) at 27 \(^\circ\) C and 10\(^5\) Nm\(^{-2}\) pressure. Its specific heat capacity at constant volume is 312 J/kg/K. Find the ratio of the specific heat at constant pressure to that at constant volume.
   Solution :
   
   density, \(\rho\) = 1.6 1.6 kgm\(^{-3}\)
   temperature, T = 300 K
   pressure, P = 10\(^5\) Nm\(^{-2}\)
   specific heat capacity at constant volume, \(c_v\) = 312 J/kg/K
   \(\gamma\) = ?
   
   From ideal gas equation,
   \[\begin{align*} P&=\rho r T\\ r&=\frac{P}{\rho T}\\ &=\frac{10^5}{1.6 \times 300}\\ r&=208.33 \hspace{0.1cm}\text{J/kg/K} \end{align*}\] Also,
   \[\begin{align*} c_p-c_v&=r\\ c_p-312&=208.33\\ c_p&=520.33 \hspace{0.1cm}\text{J/kg/K} \end{align*}\] Then the ratio of specific heat capacities is, \[\begin{align*} &=\frac{c_p}{c_v}\\ &=\frac{520.33}{312}\\ \therefore \gamma&=1.67 \end{align*}\]



10. A monoatomic ideal gas that is initially at pressure of 1.50 \(\times\) 10\(^5\) Pa and has a volume of 0.08 cu. m. compressed adiabatically to a volume of 0.04 cu.m. (a) What is the final pressure? (b) How much work is done by the gas? (c) What is the ratio of the final temperature of the gas to its initial temperature?
    Solution :
    
    For monoatomic gas, \(\gamma\) = 1.67
    initial pressure, \(P_1\) = 1.50 \(\times\) 10\(^5\) Pa
    initial volume, \(V_1\) = 0.08 m\(^3\)
    final volume, \(V_2\) = 0.04 m\(^3\)
    (a) final pressure, \(P_2\) = ?
    (b) work done by the gas, W = ?
    (c) ratio of final temperature to initial temperature, \(\frac{T_2}{T_1}\) = ?
    
    For adiabatic compression,
    
    (a)
    \[\begin{align*} P_1 V_1^\gamma &= P_2 V_2^\gamma\\ P_2&=P_1\times \left(\frac{V_1}{V_2}\right)^\gamma\\ &=1.50 \times 10^5 \times \left(\frac{0.08}{0.04}\right)^{1.67}\\ \therefore P_2&=477321.89 \hspace{0.1cm}\text{Pa} \end{align*}\] (b)
    \[\begin{align*} W&=\frac{1}{\gamma-1}(P_1V_1-P_2V_2)\\ &=\frac{1}{0.67}\times (1.5 \times 10^5 \times 0.08-477321.89 \times 0.04)\\ &=-10586.38 \hspace{0.1cm}\text{J} \end{align*}\] The negative sign indicates that the work is done on the gas.
    (c)
    \[\begin{align*} T_1V_1^{\gamma-1}&=T_2 V_2^{\gamma-1}\\ \frac{T_2}{T_1}&= \left(\frac{V_1}{V_2}\right)^{0.67}\\ &=\left(\frac{0.08}{0.04}\right)^{0.67}\\ \therefore \frac{T_2}{T_1}&=1.59 \end{align*}\]