Question

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?

Answer 1

a) Work done by gas = -15.584 KJ

b) Energy transferred = 15.584 KJ

Step-by-step 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((V_2)/(V_1))`

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

substituting the values we get,

`W=9.75* 8.31* 300.2ln((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`