Question

If $q=10454J$ for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of $1.0bar$ to a final pressure of $0.10bar$ at a constant temperature. Calculate that constant temperature.

• A. $300K$
• B. $273K$
• C. $546K$
• D. $600K$

Answer 1

The correct option is C $546K$

From the first law of thermodynamics we know,
$△U=q+w$
As it is an isothermal process the temperature remains constant,
$△U=0$,
which means,
$q=-w$
so, work done for a reversible isothermal process $=-10454J$

Also, work done,$\left(w\right)$ for a reversible isothermal process is given by
$w_{\text{rev}}=-2.303\times n\times R\times T\text{log}\frac{P_1}{P_2}$

Given
$P_1=initialpressure=1.0bar$
$P_2=finalpressure=0.1bar$

So, putting the values we get;
$-10454=-2.303\times 1\times 8.314\times T\times \text{log}\frac{1}{0.1}$
Solving this, we get
$T=546.18K$