4 mol of ideal gas undergoes a reversible and isothermal expansion from volume of 5L to 20L at 27 degree Celsius calculate the work done by the gas
Answers
Answer:
here you go.
Explanation:
condition i.e. at S.T.P
Therefore,
{V}_{i} = 22.4 \; {dm}^{3}V
i
=22.4dm
3
{V}_{f} = ?V
f
=?
As given that the expansion is isothermal and reversible
\therefore \Delta{U} = 0∴ΔU=0
Now from first law of thermodynamics,
\Delta{U} = q + wΔU=q+w
\because \Delta{U} = 0∵ΔU=0
\therefore q = – w∴q=–w
Given that the heat is absorbed.
\therefore q = 1000 \; cal∴q=1000cal
\Rightarrow w = -q = -1000 \; cal⇒w=−q=−1000cal
Now,
Work done in a reversible isothermal expansion is given by-
w = - nRT \ln{\left( \cfrac{{V}_{f}}{{V}_{i}} \right)}w=−nRTln(
V
i
V
f
)
Given:-
T = 0 ℃ = 273 \; KT=0 ℃=273K
n = 1 \text{ mol}n=1 mol
\therefore 1000 = - nRT \ln{\left( \cfrac{{V}_{f}}{{V}_{i}} \right)}∴1000=−nRTln(
V
i
V
f
)
\Rightarrow 1000 = - 1 \times 2.303 \times 2 \times 273 \times \log{\left( \cfrac{{V}_{f}}{22.4} \right)}⇒1000=−1×2.303×2×273×log(
22.4
V
f
)
\Rightarrow \log{\left( \cfrac{{V}_{f}}{22.4} \right)} = - \cfrac{1000}{2.303 \times 273 \times 2}⇒log(
22.4
V
f
)=−
2.303×273×2
1000
\Rightarrow \log{\left( \cfrac{{V}_{f}}{22.4} \right)} = -0.8⇒log(
22.4
V
f
)=−0.8
\Rightarrow {V}_{f} = 3.5 \; {dm}^{3}⇒V
f
=3.5dm
3
Hence the final volume will be 3.5 \; {dm}^{3}3.5dm