Question

At ${27}^o$C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm.
Choose the correct option from the following:

• A. Change in internal energy is positive
• B. Heat is negative
• C. Work done is - 965.84 cal
• D. All are incorrect

Answer 1

Answer: (B)

Heat is negative

Work done in isothermal reversible process is

$w=-2.303nRTlo{g}_{10}\frac{p_1}{p_2}$

Given, $n=1,R=2cal{K}^{-1}mo{l}^{-1}$

$T=(27+273)K=300K$

$p_1=2atm,p_2=10atm$

$\therefore$ $w=-2.303\times 1\times 2\times 300lo{g}_{10}\frac{2}{10}$

$w=+965.54cal$

For isothermal change, $\Delta U=0$

Now, from first law of thermodynamics

$q=\Delta U-W=0-965.84cal$

$q=-965.84cal$