Question 73: 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

Choose the correct answer: