Question

An ideal gas contained in a piston–cylinder device undergoes an isothermal compression process which begins with an initial pressure and volume of 120 kPa and 0.6 m3, respectively. During the process, there is a heat transfer of 60 kJ from the ideal gas to the surroundings. Determine the volume and the pressure at the end of the process. The properties of air are R = 0.287 kJ/kg·K and cv = 0.718 kJ/kg·K.

Answer 1

Explanation:

As it is given that no change in the internal energy is taking place so, transfer of heat will be equal to the work done by the system. Hence, we will obtain the final volume as follows.

              W = $-\int_{V_{1}}^{V_{2}}P dV$

                   = $-mRT \int_{V_{1}}^{V_{2}} \frac{1}{V} dV$

                   = $-P_{1}V_{1} ln \frac{V_{2}}{V_{1}}$

           $V_{2} = V_{1} e^{\frac{-W}{P_{1}V_{1}}}$

                        = $0.6 \times e^{\frac{-60}{100.06}} m^{3}$

                        = 0.221 $m^{3}$

Now, using the volume ratio we will determine the volume ratio as follows.

                 $P_{2} = P_{1} \times \frac{V_{1}}{V_{2}}$

                             = $100 \times \frac{0.6}{0.221} kPa$

                             = 271.5 kPa

thus, we can conclude that volume is 0.221 $m^{3}$ and pressure is 271.5 kPa at the end of the process.

Answer 2

Answer:

The volume and the pressure at the end of the process is 0.1497 m³ and 480.961 kPa.

Explanation:

Given that,

Initial pressure = 120 kPa

Volume = 0.6 m³

Heat = 60 kJ

We need to calculate the volume

Using formula of work done

$W=P_{i}V_{i}\ ln(\dfrac{V_{f}}{V_{i}})$

Put the value into the formula

$-100\times10^{3}=120\times10^{3}\times0.6\ ln(\dfrac{V_{f}}{0.6})$

$-100=72\ ln(\dfrac{V_{f}}{0.6})$

$\dfrac{-100}{72}=ln(\dfrac{V_{f}}{0.6})$

$e^{-1.388}=\dfrac{V_{f}}{0.6}$

$V_{f}=e^{-1.388}\times0.6$

$V_{f}=0.1497\ m^3$

We need to calculate the pressure

Using formula of pressure

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

Put the value into the formula

$P_{2}=\dfrac{120\times10^{3}\times0.6}{0.1497}$

$P_{2}=480961.9\ Pa$

$P_{2}=480.961\ kPa$

Hence, The volume and the pressure at the end of the process is 0.1497 m³ and 480.961 kPa.