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A container holds 0.1 \mathrm{~kg} of an ideal gas initially at a pressure of 5 \mathrm{~atm} and a volume of 0.02 \mathrm{~m}^3. The gas is subjected to a polytropic process described by the equation P V^n= constant, where n is a constant. The final volume of the gas is 0.1 \mathrm{~m}^3, and the final pressure is 3 \mathrm{~atm}. Given that the heat capacity ratio for the gas is 1.4 , calculate the value of n.
 

Option: 1

1.405 \mathrm{k}


Option: 2

1.70 \mathrm{k}


Option: 3

20.30 \mathrm{k}


Option: 4

23.20 \mathrm{k}


Answers (1)

best_answer

For a polytropic process, the relationship between pressure (P) and volume (V) of an ideal gas is given by:
\mathrm{P V^n=\text { constant } }
During the process, the total work done by the gas is given by:
\mathrm{W=-\frac{\text { constant }}{1-n}\left(V_f^{1-n}-V_i^{1-n}\right) }

Given that\mathrm{ V_i=0.02 \mathrm{~m}^3, V_f=0.1 \mathrm{~m}^3, P_i=5 \mathrm{~atm}, and P_f=3 \mathrm{~atm}, } we can rewrite the equation using the ideal gas law:
\mathrm{\frac{P_i V_i}{n R}=\frac{P_f V_f}{n R} }
Solving for n :
\mathrm{n=\frac{\ln \left(\frac{P_i V_i}{P_f V_f}\right)}{\ln \left(\frac{V_i}{V_f}\right)} }
Substitute the given values and the heat capacity ratio \mathrm{ \gamma }=1.4 :
\mathrm{n=\frac{\ln \left(\frac{5 \mathrm{~atm} \times 0.02 \mathrm{~m}^3}{3 \mathrm{~atm} \times 0 . \mathrm{m}^3}\right)}{\ln \left(\frac{0.02 \mathrm{~m}^3}{0.1 \mathrm{~m}^3}\right)}=\frac{\ln (0.3333)}{\ln (0.2)} \approx 1.405 }
Therefore, the value of n for the polytropic process is approximately 1.405 .

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shivangi.bhatnagar

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