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Bond [772]
3 years ago
11

at a constant temperature a gas is compressed from 4L to 1L. At 1L, the pressure of the gas is 98.2 kPa. What is the original pr

essure of this gas?​
Chemistry
1 answer:
Elis [28]3 years ago
8 0

Answer: 24.6kpa

Explanation:

This is a classic case where Boyle's Law can be applied.

The equation for Boyle's Law is given as: P1V1 = P2V2

Where P1 = initial pressure

P2 = final pressure

V1 = initial volume

V2 = final volume

From the question, P1 = ?, P2 = 98.2kpa , V1 =4L , V2 =1L

P1V1 = P2V2

P1 x 4L = 98.2 kpa x 1L

Make P1 subject of formula we then have:

P1 = 98.2kpa X 1L / 4L

= 24.6kpa

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A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
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Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

PV_1=nRT

where,

P = pressure of gas = 1 atm

V_1 = volume of gas = ?

T = temperature of gas = 270^oC=273+270=543K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K

V_1=9.12L

Now we have to calculate the volume at 24°C.

PV_2=nRT

where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 9.12 L

V_2 = final volume = 4.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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