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A gas at atmospheric pressure is contained in a cylinder of volume 80 litre. When it is compressed adiabatically to 20 litre its pressure rises to $\mathrm{7 \mathrm{~atm}}$ . What will be the ratio of specific heats of the gas

Option: 1
$1.33$

Option: 2
$1.4$

Option: 3
$1.67$

Option: 4
$1.5$

Answers (1)

best_answer

$\mathrm{P}_{\mathrm{i}}=1 \mathrm{~atm}=1 \times 10^5 \mathrm{~N} / \mathrm{m}^2$

$\mathrm{ V_i=80 \times 10^{-3} \mathrm{~m}^3 }$

$\mathrm{ V_f=20 \times 10^{-3} \mathrm{~m}^3 }$

$\mathrm{ P_f=7 \text { atm }=7 \times 10^5 \mathrm{~N} / \mathrm{m}^2 }$

$\mathrm{ P_{i V_i}=P_f V_f^\gamma }$

$\mathrm{ 1 \times 10^5 \times\left(80 \times 10^{-3}\right)^\gamma=\left(7 \times 10^5\right)\left(20 \times 10^{-3}\right)^\gamma }$

$\mathrm{\left(\frac{80 \times 10^{-3}}{20 \times 10^{-3}}\right)^\gamma=7 }$

$\mathrm{ (4)^\gamma=7 }$

$\mathrm{ \gamma \log 4=\log 7 }$

$\mathrm{ \gamma=\frac{\log 7}{\log 4}=1.40 }$

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shivangi.shekhar

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