One mole of oxygen gas having a volume equal to 22.4 litres at $0^{\circ} C$ and 1 atm pressure is compressed isothermally so that it's volume reduces to 11.2 litres. What is the work done in this process?

Open in App

Solution

Work done in an isothermal process is given by
$W = n R T l n (\frac{V_{2}}{V_{1}})$
$= 1 \times 8.314 \times 273 l n (\frac{11.2}{22.4})$
$= 1 \times 8.314 \times 273 l n (\frac{1}{2})$
$= 8.314 \times 273 \times - 0.69$
$= 1566.108 J$

Suggest Corrections

Similar questions

Q. One mole of

O_{2}

gas having a volume equal to

22.4

litres at

0^{\circ} C

and 1 atmospheric pressure is compressed isothermally so that its volume reduces to

11.2

litres. The work done in this process is

One mole of a gas having a volume equal to 22.4 litres and at 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is

Q. One mole of

O_{2}

gas having a volume equal to 22.4 litres at

0^{\circ}

C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is

Q. One mole of

O_{2}

gas having a volume equal to 22.4 liters at

0^{o} C

and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 liters. The work done in this process is then

Q. Assertion :8 grams of methane occupies 11.2 litres of volume at 273

K

and 1

a t m

pressure. Reason: One mole of any gas at

S T P

occupies 22.4 litres of volume.

Join BYJU'S Learning Program

One mole of oxygen gas having a volume equal to 22.4 litres at 0∘C and 1 atm pressure is compressed isothermally so that it's volume reduces to 11.2 litres. What is the work done in this process?

Work done in an isothermal process is given by W=nRT ln(V2V1) =1×8.314×273 ln(11.222.4) =1×8.314×273 ln(12) =8.314×273×−0.69 =1566.108 J

One mole of oxygen gas having a volume equal to 22.4 litres at $0^{\circ} C$ and 1 atm pressure is compressed isothermally so that it's volume reduces to 11.2 litres. What is the work done in this process?

Work done in an isothermal process is given by
$W = n R T l n (\frac{V_{2}}{V_{1}})$
$= 1 \times 8.314 \times 273 l n (\frac{11.2}{22.4})$
$= 1 \times 8.314 \times 273 l n (\frac{1}{2})$
$= 8.314 \times 273 \times - 0.69$
$= 1566.108 J$