0.180 mol of an ideal gas, at 27.17 °C, is expanded isothermally from 3.06 L to

Question

0.180 mol of an ideal gas, at 27.17 °C, is expanded isothermally from 3.06 L to 10.28 L.

1. What is the initial pressure of the gas, in atm?

2. What is the final pressure of the gas, in atm?

3. How much work is done on the gas, (in J), if the expansion is carried out by changing the volume instantaneously from 3.06 L to 10.28 L, followed by allowing the pressure to reach its final value at the constant final volume?

4. How much work is done on the gas, (in J), if the expansion is done under isothermal, reversible conditions?

Explanation / Answer

according to ideal gas equation

PV = nRT

where P =pressure, V= volume, n=mole = 0.180, R= gas constant = 0.082 lit-atm/mol/K, T = temperature = 300.17 K

1. P *3.06 = 0.18*0.082*300.17

or, P = 1.45 atm

2. similarly,  P *10.28 = 0.18*0.082*300.17

or, P = 0.43 atm

3. The volume has been changed instantaneously . so, it is an free expansion means external pressure is zero,

therefore, work done will be zero, ( work done = - external pressure * change in volume)

4. it is an isothermal, reversible conditions.

so, work done will be

W=nRTln(V2/V1) = - 0.18*8.314 J/mol.K* 300.17 K * ln (10.28 L/3.06 L) = 544.34 J (ans)

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