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3 years ago

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An ideal gas undergoes isothermal compression from an initial volume of 5.39 m3 to a final volume of 2.84 m3. There is 9.75 mol

of the gas, and its temperature is 27.2°C. (a) How much work is done by the gas? (b) How much energy is transferred as heat between the gas and its environment?

1 answer:

Anika [276]3 years ago

6 0

Answer:

a) Work done by gas = -15.584 KJ

b) Energy transferred = 15.584 KJ

Explanation:

Given:

Initial volume of the gas, V₁ = 5.39 m³

Final volume, V₂ = 2.84 m³

Temperature, T = 27.2°C = 273+27.2 = 300.2K

Number of moles, n = 9.75 mol

a) Work done for the given isothermal process is

$W=nRTln(\frac{V_2}{V_1})$

where, R is the ideal gas constant = 8.31 J/mol.K

substituting the values we get,

$W=9.75\times 8.31\times 300.2ln(\frac{2.84}{5.39})$

or

$W=-15584.72 J = -15.584 KJ$

b) From the first law of thermodynamics

change in internal energy = Heat added - Work done

or, $\Delta Q = \Delta U - W$

Now for the isothermal process, ΔU = 0

thus,

$\Delta Q = - W$

or

$\Delta Q = - (-15.584 KJ) = 15.584 KJ$

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