Question

3.111 Air contained in a piston-cylinder assembly contains air, initially at 2 bar, 300 K and a volume of 2 m^3. The air undergoes a process to a state where pressure is 1 bar, during which the pressure-volume relationship is PV=constant. Assuming ideal gas behavior for air, determine the mass of the air, in kg and the work and heat transfer, each in KJ.

Answer 1

Concepts and Reason

An ideal gas is a theoretical gas consisting of particles that move randomly and collide elastically with each other. All ideal gases obey the ideal gas law. Air in most cases, can be considered to be an ideal gas. When PV = constant, it indicates that the temperature is constant during that process (Boyle’s Law). It is an isothermal process.

In a piston-cylinder assembly containing air, the piston compresses the air inside the cylinder causing the air to ignite and do work on the piston in return. It follows the law of conservation of energy.

Fundamentals

The ideal gas law is given by,

$PV = mRT$

Here, the pressure is P, the volume is V, the mass of air is m, the specific gas constant is R, and the temperature is T.

The work done in an isothermal process is given by,

$\begin{array}{l}\\W = mRT\ln \left( {\frac{{{P_1}}}{{{P_2}}}} \right)\\\\{\rm{or}}\\\\W = mRT\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\\\end{array}$

Here, the pressures at points 1 and 2 are ${P_1}$ and ${P_2}$ respectively. The volumes at points 1 and 2 are ${V_1}$ and ${V_2}$ respectively.

The change in internal energy in an isothermal process is zero.

From the first law of thermodynamics, due to the internal energy being zero in an isothermal process, the heat transferred is numerically equal to the work done.

$Q = W$

Here, the heat transferred is Q and the work done is W.

Calculate the mass of air using the formula,

${P_1}{V_1} = mRT$

Here, the pressure at state 1 is ${P_1}$ , the volume at state 1 is ${V_1}$ , the mass of air is m, the specific gas constant is R, and the temperature is T.

Substitute $2 \times {10^5}{\rm{ Pa}}$ for ${P_1}$ , $2{\rm{ }}{{\rm{m}}^3}$ for ${V_1}$ , $287{\rm{ J/kgK}}$ for R, and 300 K for T.

$\begin{array}{l}\\2 \times {10^5} \times 2 = m \times 287 \times 300\\\\m = 4.646{\rm{ kg}}\\\end{array}$

Calculate the work done in an isothermal process. Isothermal process indicates that temperature is the same at states 1 and 2.

$W = mRT\ln \left( {\frac{{{P_1}}}{{{P_2}}}} \right)$

Here, the pressures at points 1 and 2 are ${P_1}$ and ${P_2}$ respectively.

Substitute 2 bar for ${P_1}$ , 1 bar for ${P_2}$ , $0.287{\rm{ kJ/kgK}}$ for R, 4.646 kg for m, and 300 K for T.

$\begin{array}{c}\\W = 4.646 \times 0.287 \times 300 \times \ln \left( {\frac{2}{1}} \right)\\\\ = 400.02 \times \ln 2\\\\ = 277.27{\rm{ kJ}}\\\end{array}$

The change in internal energy in an isothermal process is zero.

From the first law of thermodynamics, due to the internal energy being zero in an isothermal process, the heat transferred is numerically equal to the work done.

Calculate the heat transfer.

$Q = W$

Substitute 277.27 kJ for W.

$Q = 277.27{\rm{ kJ}}$

Ans:

The mass of the air is 4.646 kg.

The work done is 277.27 kJ.

The heat transfer is 277.27 kJ.