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?​

Answer 1

Explanation:

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.

Initial pressure inside the cylinder = P1

Final pressure inside the cylinder = P2

Initial volume inside the cylinder = V1

Final volume inside the cylinder = V2

Ratio of specific heats, γ = 1.4

For an adiabatic process, we have:

P1V1γ = P2V2γ

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

∴ V2 = V1/2

P1V1γ = P2(V1/2)γ

P2/P1 = V1γ / (V1/2)γ

= 2γ = 21.4 = 2.639

Hence, the pressure increases by a factor of 2.639.

Answer 2

Answer:

Explanation:Now here the cylinder or the container containing the gas and piston mounted are both insulated so no transfer of heat will take place either from system to surrounding and vice versa , so the process will be adiabatic as the heat supplied to or from the system

ΔQ = 0

Now the system consists of 3 Mole Hydrogen gas kept at standard temperature and pressure in an insulated container

Now here the gas is compressed to half of its volume, we have to calculate factor of change of pressure of gas when volume is reduced to half

We know for an adiabatic process we have the relation

Where P1 and P2 are initial and final pressures of the system respectively, V1 and V2 are initial and final volume of the system respectively, γ is the specific heat ratio of the gas since, H2 is a diatomic gas, for a diatomic gas

γ = 1.4

now let the initial volume of the system be

V1 = V

Now here the gas is compressed to half of its volume so final volume of the system is

V2 = V/2

Let initial and final pressures of the system be P1 and P2 respectively

Now applying the relation

We have  p1v1^y =p2v2^y=p1v^1.4=p2 [v/2}^1.4

Or  p2/p1=v^1.4*2^1.4/v^1.4=2^1.4=p2/p1=2.639

i.e.  

so we can say that the final pressure is 2.639 times initial pressure or the pressure increased by a factor of 2.639

here ^ denotes that the number after this symbol should be up

hope it helps!