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# n moles of an ideal gas, xs, having a bent geometry, is heated to a high temperature t from initial temperature ti in an piston with a movable plug at constant pressure. the external pressure is then lowered and the gas is allowed to expand isothermally until the volume increases by 22%. develop an expression for the total entropy change. explain and justify your approach.

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get n moles of an ideal gas, xs, having a bent geometry, is heated to a high temperature t from initial temperature ti in an piston with a movable plug at constant pressure. the external pressure is then lowered and the gas is allowed to expand isothermally until the volume increases by 22%. develop an expression for the total entropy change. explain and justify your approach. from EN Bilgi.

## n moles of an ideal gas, xs, having a bent geometry, is heated to a high temperature t from initial temperature ti in an piston with a movable plug ..

question is : n moles of an ideal gas, xs, having a bent geometry, is heated to a ... the total entropy change. explain and justify your approach.

## n moles of an ideal gas, xs, having a bent geometry, is heated to a high temperature t from initial temperature ti in an piston with a movable plug ..

n moles of an ideal gas, xs, having a bent geometry, is heated to a high temperature t from initial temperature ti in an piston with a movable plug .. 1 Answer

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## A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Click here👆to get an answer to your question ✍️ A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume? Question

## A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.

Medium Open in App Solution Verified by Toppr

Initial pressure inside the cylinder =P

1 ​

Final pressure inside the cylinder =P

2 ​

Initial volume inside the cylinder =V

1 ​

Final volume inside the cylinder =V

2 ​

Ratio of specific heats, γ=

C V ​ C P ​ ​ =1.4

For an adiabatic process, we have:

P 1 ​ V 1 γ ​ =P 2 ​ V 2 γ ​

The final volume is compressed to half of its initial volume.

∴V 2 ​ =V 1 ​ /2 P 1 ​ V 1 γ ​ =P 2 ​ (V 1 ​ /2) γ P 2 ​ /P 1 ​ =V 1 γ ​ /(V 1 ​ /2) γ =2 1.4 =2.639

Hence, the pressure increases by a factor of 2.639.

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## Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movable piston, exposed to the atmosphere, at an initial temperature T(0). The gas is slowly heated so that its volume becomes four times the initial value. The work done by gas is

d. Since the gas is slowly heated, it remains in equilibrium (more or less) with the atmosphere, i.e., the process takes place at a constant pressure. Now, from PV = nRT, PdV = nRdT or, P Delta V = nR Delta T But P Delta V is the work done by the gas. So, Delta W = nR Delta T = (2 mol) (R ) (4 T(0) - T(0)) = 6 RT(0) [From Eq. (i) Delta V prop Delta T, i.e., if Delta V = 3V(0), Delta T = 3 T(0)] Home > English > Class 11 > Physics > Chapter >

Kinetic Theory Of Gases And First Law Of Thermodynamics

>

Two moles of an ideal gas is c...

Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movable piston, exposed to the atmosphere, at an initial temperature

T 0 T0

. The gas is slowly heated so that its volume becomes four times the initial value. The work done by gas is

Updated On: 27-06-2022

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Text Solution Open Answer in App A zero B 2R T 0 2RT0 C 4R T 0 4RT0 D 6R T 0 6RT0 Answer

Solution

d. Since the gas is slowly heated, it remains in equilibrium (more or less) with the atmosphere, i.e., the process takes place at a constant pressure.

Now, from PV=nRT,PdV=nRdT PV=nRT,PdV=nRdT or, PΔV=nRΔT PΔV=nRΔT But PΔV PΔV

is the work done by the gas.

So, ΔW=nRΔT=(2mol)(R)(4 T 0 − T 0 )=6R T 0

ΔW=nRΔT=(2mol)(R)(4T0-T0)=6RT0

[From Eq. (i) ΔV∝ΔT ΔV∝ΔT , i.e., if ΔV=3 V 0 ,ΔT=3 T 0 ΔV=3V0,ΔT=3T0 ] Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 CHAAYA PRAKASHINI (BENGALI) HC VERMA ENGLISH XII BOARD PREVIOUS YEAR PAPER ENGLISH NCERT EXEMPLAR ENGLISH KUMAR PRAKASHAN MODERN PUBLICATION FULL MARKS CBSE COMPLEMENTARY MATERIAL SUBHASH PUBLICATION VGS PUBLICATION-BRILLIANT SURA PUBLICATION CHHAYA PUBLICATION MARVEL PUBLICATION XII BOARDS PREVIOUS YEAR ARIHANT NCERT EXEMPLAR HC VERMA SUNIL BATRA (41 YEARS IITJEE PHYSICS) DC PANDEY CENGAGE PHYSICS A2Z  IE IRODOV, LA SENA & SS KROTOV CP SINGH NCERT FINGERTIPS Source : www.doubtnut.com

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James 1 month ago

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