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A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 370 C. As it does so ,it absorbs 208 J of heat. The values of q and w for the process will be :

(R= 8.314 J/mol K) (In 7.5 = 2.01)

Option: 1

q = +\: 208 \: J,\: w = +\: 208\: J


Option: 2

q = +\: 208 \: J,\: w = -\: 208\: J


Option: 3

q = -\: 208 \: J,\: w = -\: 208\: J


Option: 4

q = -\: 208 \: J,\: w = +\: 208\: J


Answers (1)

best_answer

The process is isothermal reversible expansion

Hence dT = 0

\DeltaU = 0

Now, work is given by the formula

\mathrm{W = -n\times R\times T\times ln(\frac{V_2}{V_1})}

\mathrm{W = - 0.04\times 8.314\times 310\times ln(\frac{375}{50})= -208\ J}

Now, using 1st law of thermodynamics

\mathrm{Q + w = \Delta U}

\therefore \mathrm{Q =-w = 208\ J}

Hence, the correct answer is Option (2)

Posted by

Rakesh

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