The one mole of an ideal gas does 2900J of work on surroundings as it expands isothermally. The initial volume of a gas is 8.0 L.
FIrst find out the temperature of a gas using ideal gas law which is
PV=nRT
For ideal gas, n=1 and R=0.08206 Latm/molK, V=25.0L, P=1atm
Rearrange the formula for T
T=PV/nR
Plug all values in the formula
T=(1.00atm×25.0L×molK/1mol×0.08206Latm)
T=25.00 K/0.08206
T=304.65K
The work done for isothermal condition is calculated as
W=2.303nRTlog(V₂/V₁)
Where V₂=25.0L, R=8.314 J/molK, n=1 mol, T=304.65K, V₁=?, W=2900J
Plug all values in the formula
2900J=2.303×1mol×(8.314J/molK)×304.65K×log(25/V₁)
2900J=5833.1768J×log(25/V₁)
log(25/V₁)=(2900 J/5833.1768 J)
log(25/V₁)=0.49716
Take antilog on both side
25/V₁=10^0.49716
25/V₁=3.14167
V₁=(25/3.14167)
V₁=7.95755 L
V₁=8.0 L
Therefore, the initial volume of a gas is 8.0 L.
To know more about isothermal expansion
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