Question

3 mole of an ideal gas at 1.5 ATM and 25 degree Celsius expand isothermally and reversible manner to twice its original volume against an external pressure of 1 ATM calculate the work done.( R = 8.314 JK^-1 mol^-1)

Answer 1

Work is done by the system of magnitude 5154.56 J.

Explanation:

Given that:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25.0 + 273.15) K = 298.15 K

The expression for the work done is:

$W=-nRT \ln \left( \dfrac{V_2}{V_1} \right)$

Where,  

n is the number of moles = 3 moles

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol  

T is the temperature  

V₁ is the initial volume

V₂ is the final volume

The final volume is double of the original, So, V₂ = 2V₁

Given that:

T = 298.15 K

The final volume is double of the original, So,

Applying in the equation as:

$W=-3\ moles\times 8.314\ J/Kmol\times 298.15\ K \ln \left( \dfrac{2\times V_1}{V_1} \right)$

$W=-3\times 8.314\times 298.15 \ln 2\ J=-7436.4573\ln \left(2\right)\ J=-0.69314 \times \:7436.4573$

$W=-5154.56\ J$ (Work done by system)

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Answer 2

Answer:5152.89

Explanation:3 mol at 25°C

Temp = 273 + 25 = 298K

V¹ = V =

V² = 2*V =

W = -2.303 nRT log(V²/)V¹)

W = 5152.89J