A horizontal cylinder with adiabatic walls is closed at both ends and is divided into two parts by a frictionless piston which is insulating-Initially- the value of pressure and temperature of the ideal gas on each side of the cylinder are V0 - P0 and T0- respectively-A heating coil in the right hand part is used to slowly heat the gas on that side until the pressure in both parts reaches 64 P027-Take CPCv - - - 1-5-V0 - 16 m3 - T0 - 324 K- P0 - 3 - 105 Pa-Column I represents the physical parameters of the gas and Column II give their corresponding values- Match Column I with Column II- Column I Column II- i- Final volume of left hand side part in m3 a- 432- ii- Final temperature left hand side part in K b- 9- iii- Final temperature right hand side part in K c- 1104- iv- Work done in kJ on gas in left hand side part d- 3200- -

I → b.; ii → a.; iii → c.; iv → d. The compression in the left hand side is adiabatic, P_0 V^γ_0 = P_1 V^γ_1 V_1 = V_0 ( P_0P_1)^23 = V_0 ( 2764)^23 = 9V_ ...

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Byju's Answer

Standard XIII

Physics

Thermodynamic Processes

A horizontal ...

Question

A horizontal cylinder with adiabatic walls is closed at both ends and is divided into two parts by a frictionless piston which is insulating.
Initially, the value of pressure and temperature of the ideal gas on each side of the cylinder are $V_{0}, P_{0}$ and $T_{0}$ , respectively.
A heating coil in the right hand part is used to slowly heat the gas on that side until the pressure in both parts reaches $\frac{64 P_{0}}{27}$ .
Take $\frac{C_{P}}{C_{v}} = γ = 1.5,$ $V_{0} = 16 m^{3}, T_{0} = 324 K, P_{0} = 3 \times 10^{5} Pa$ .

Column I represents the physical parameters of the gas and Column II give their corresponding values. Match Column I with Column II.

$\begin{matrix} Column I & Column II i. Final volume of left hand side part ({in m}^{3}) & a . 432 ii. Final temperature left hand side part (in K) & b . 9 iii. Final temperature right hand side part (in K) & c . 1104 iv. Work done (in kJ) on gas in left hand side part & d . 3200 \end{matrix}$

i - b, i i - c, i i i - a, i v - d

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i - c, i i - b, i i i - a, i v - d

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i - b, i i - a, i i i - c, i v - d

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i - b, i i - c, i i i - d, i v - a

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Solution

The correct option is C $i - b, i i - a, i i i - c, i v - d$
$i \to b .; i i \to a .; i i i \to c .; i v \to d .$

The compression in the left - hand side is adiabatic,
$P_{0} V_{0}^{γ} = P_{1} V_{1}^{γ}$

$V_{1} = V_{0} {(\frac{P_{0}}{P_{1}})}^{\frac{2}{3}} = V_{0} {(\frac{27}{64})}^{\frac{2}{3}} = \frac{9 V_{0}}{16} = 9 m^{3}$
Also
$\frac{P_{0} V_{0}}{T_{0}} = \frac{P_{1} V_{1}}{T_{1}}$
$T_{1} = \frac{P_{1} V_{1}}{P_{0} V_{0}} = \frac{4 T_{0}}{3} = 432 K$

For right hand side part
$P_{2} = \frac{64 P_{0}}{27}$ and $V_{2} = 2 V_{0} - V_{1} = 23 m^{3}$
$\frac{P_{0} V_{0}}{T_{0}} = \frac{P_{2} V_{2}}{T_{2}}$

$T_{2} = \frac{92 T_{0}}{27} = 1104 K$
Work done on the left - hand side gas is
$W = \frac{P_{1} V_{1} - P_{0} V_{0}}{γ - 1} = \frac{(\frac{64}{27} \times \frac{9}{16} - 1) P_{0} V_{0}}{\frac{3}{2} - 1}$
$= \frac{2}{3} P_{0} V_{0} = 3200 kJ$

Why this question ?

Key Concept: Thermal insulation prevents heat transfer. So,gas in left hand side of the cylinder undergoes adiabatic process.

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