Question 77: Thermodynamics & KTG

Question

There are two identical containers C$_1$ and C$_2$ containing to identical gases. Gas in C$_1$ is reduced to half of its original volume adiabatically, while the gas in container C$_2$ is also reduced to half of its initial volume isothermally. Find the ratio of final pressure in these containers. ($\gamma$ be the adiabatic constant).

Accepted Answer

Since, container are identical, so initial volume and pressure will be same for identical gases.
For container C$_1$,
${p_1}V_1^\gamma  = {p_2}V_2^2$ (adiabatic)
${p_2} = {\left( {{{{V_1}} \over {{V_2}}}} \right)^\gamma }{p_1} = {\left( {{{{V_1}} \over {{V_1}/2}}} \right)^\gamma }{p_1} = {2^\gamma }{p_1}$ ..... (i)
For container C$_2$,
${p_1}{V_1} = {p_2}'{V_2}$ (isothermal)
${p_2}' = \left( {{{{V_1}} \over {{V_2}}}} \right){p_1} = \left( {{{{V_1}} \over {{V_1}/2}}} \right){p_1} = 2{p_1}$ ..... (ii)
From Eqs. (i) and (ii), we get
${{{p_2}} \over {{p_2}'}} = {{{2^\gamma }{p_1}} \over {2{p_1}}} = {2^{\gamma  - 1}}:1$

Answer

2 : 1

Answer

1 : 2

Answer

$2^\gamma : 1$

Answer

$2^{\gamma-1}:1$

Thermodynamics & KTG

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