Question

a quantity of of gas is compressed according to pv^1.25 = constant.the initial temperature and pressure of the gas is 15°c and 1 bar respectively.find the work done in compressing 1kg of air at 3 bar and the heat rejected through the wall of the cylinder (take Gama = 1.4 for air)​

Answer 1

Answer:

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Answer 2

Given:

A quantity of of gas is compressed according to pv^1.25 = constant. the initial temperature and pressure of the gas is 15°c and 1 bar respectively.

To Find:

Find the work done in compressing 1kg of air at 3 bar and the heat rejected through the wall of the cylinder (take Gama = 1.4 for air)​.

Solution:

Given;

mass of gas, m=1 kg

Initial pressure, $P_{1}=1 \ bar=10^5Pa$

Final pressure, $P_{2}=3 \ bar=3\times10^5 Pa$

Initial temperature, $T_{1}=15+273=288 K$

given that - $\gamma = 1.4$ implies, $n=\gamma=1.4 \ for \ constant \ change \ in \ temperature.$

since, we know that -

$\dfrac{T_{2}}{T_{1}}=\left (\dfrac{P_{2}}{P_{1}}\right)^{1-\dfrac{1}{n}}$

therefore,

$T_{2}=288\left (\dfrac{3\times10^5}{10^5}\right)^{1-\frac{1}{1.4}} \\\\=288(3)^{1-0.72}\\\\=288\times1.36\\\\=391.68 K$

take, R = 277.13 J/kg/K.

Now, work done in compression is given by;

$W=\dfrac{P_{1}V_{1}-P_{2}V_{2}}{n-1}=\dfrac{mR(T_{1}-T_{2})}{n-1}\\\\=\dfrac{1\times 277.13\times(288-391.68)}{1.4-1}\\\\=-7182.096 \ J\\\\=-7.182 \ kJ$