Q32 / 180 · JIPMER 2016 · Chemistry

At $27^{\circ} \mathrm{C}$, 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
  • Correct answer
    B
    Heat is negative
  • C
    Work done is -965.84 cal$
  • D
    All are incorrect

Solution

Work done in isothermal reversible process is $w=-2.303 n R T \log _{10} \frac{p_1}{p_2}$ Given, $\begin{aligned} n & =1, R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\ T & =(27+273) \mathrm{K}=300 \mathrm{~K} \\ p_1 & =2 \mathrm{~atm}, p_2=10 \mathrm{~atm}\end{aligned}$ $\begin{aligned} \therefore \quad w & =-2.303 \times 1 \times 2 \times 300 \log _{10} \frac{2}{10} \\ w & =+965.54 \mathrm{cal}\end{aligned}$ For isothermal change, $\Delta U=0$ Now, from first law of thermodynamics, $\begin{aligned} & q=\Delta U-W=0-965.84 \mathrm{cal} \\ & q=-965.84 \mathrm{cal}\end{aligned}$

Hence, the correct answer is option: 'B'.

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