Heat And Thermodynamics Ques 128
- Half-mole of an ideal monoatomic gas is heated at constant pressure of $1 atm$ from $20^{\circ} C$ to $90^{\circ} C$. Work done by gas is close to (Take, gas constant, $R=8.31 J / mol-K$ )
(Main 2019, 10 Jan II)
(a) $291 J$
(b) $581 J$
(c) $146 J$
(d) $73 J$
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Answer:
Correct Answer: 128.(a)
Solution:
Formula:
- Work done by gas during heat process at constant pressure is given by
$$ \Delta W=p \Delta V $$
Using ideal gas equation,
$$ \begin{array}{rlrl} & & p V & =n R T \\ \Rightarrow & & p \Delta V & =n R \Delta T \\ \text { So, } & \Delta W & =n R \Delta T \cdots(i) \end{array} $$
Now, it is given that, $n=\frac{1}{2}$
and
$$ \Delta T=90^{\circ} C-20^{\circ} C $$
$$ =363 K-293 K=70 K $$
and $R$ (gas constant) $=8.31 J / mol-K$
Substituting these values in Eq. (i), we get
$$ \begin{aligned} \Delta W & =\frac{1}{2} \times 8.31 \times 70=290.85 J \\ \Delta W & \simeq 291 J \end{aligned} $$
BETA
