Answer:

Heat transfer in the process is -19.6 kJ

Work done in the process is – 28 kJ

Explanation:

As we know that the process equation is given as

[tex]P_1V_1^{1.1} = P_2V_2^{1.1}[/tex]

now we know that

[tex]P_1 = 1 bar[/tex]

[tex]V_1 = 0.2 m^3[/tex]

[tex]P_2 = 4 bar[/tex]

now from above equation we have

[tex]1(0.2)^{1.1} = 4(V)^{1.1}[/tex]

[tex]V = 0.057 m^3[/tex]

now work done in this process is given as

[tex]W = \frac{P_1V_1 – P_2V_2}{N – 1}[/tex]

here we know N = 1.1

so we have

[tex]W = \frac{10^5 (0.2) – 4 \times 10^5(0.057)}{1.1 – 1}[/tex]

[tex]W = -2.8 \times 10^4 J[/tex]

[tex]W = -28 KJ[/tex]

Now we have

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]T_2 = \frac{P_2V_2 T_1}{P_1V_1}[/tex]

[tex]T_2 = \frac{4\times 10^5 (0.057) (300)}{10^5 (0.2)}[/tex]

[tex]T_2 = 342 K[/tex]

now we have

[tex]\Delta U = \frac{f}{2} n R(\Delta T)[/tex]

[tex]\Delta U = 3(P_2 V_2 – P_1 V_1)[/tex]

[tex]\Delta U = 3(4 \times 10^5(0.057) – 10^5(0.2))[/tex]

[tex]\Delta U = 8400[/tex]

so we have

[tex]Q = \Delta U + W[/tex]

[tex]Q = -28 kJ + 8.4 kJ[/tex]

[tex]Q = -19.6 kJ[/tex]

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