SATHEE: Heat and Thermodynamics 5 Question 10

Heat and Thermodynamics 5 Question 10

10. Half-mole of an ideal monoatomic gas is heated at constant pressure of $1 a t m$ from $20^{\circ} C$ to $90^{\circ} C$ . Work done by gas is close to (Take, gas constant, $R = 8.31 J / m o l - K$ )

(Main 2019, 10 Jan II)

(a) $291 J$

(b) $581 J$

(d) $73 J$

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Solution:

Work done by gas during heat process at constant pressure is given by

$Δ W = p Δ V$

Using ideal gas equation,

$\begin{aligned} p V & = n R T \\ \Rightarrow & p Δ V & = n R Δ T \\ So, & Δ W & = n R Δ T \end{aligned}$

Now, it is given that, $n = \frac{1}{2}$

and

$Δ T = 90^{\circ} C - 20^{\circ} C$

$= 363 K - 293 K = 70 K$

and $R$ (gas constant) $= 8.31 J / m o l - K$

Substituting these values in Eq. (i), we get

$\begin{aligned} Δ W & = \frac{1}{2} \times 8.31 \times 70 = 290.85 J \\ Δ W & ≃ 291 J \end{aligned}$

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Heat and Thermodynamics 5 Question 10

10. Half-mole of an ideal monoatomic gas is heated at constant pressure of 1atm from 20∘C to 90∘C. Work done by gas is close to (Take, gas constant, R=8.31J/mol−K )

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10. Half-mole of an ideal monoatomic gas is heated at constant pressure of $1 a t m$ from $20^{\circ} C$ to $90^{\circ} C$ . Work done by gas is close to (Take, gas constant, $R = 8.31 J / m o l - K$ )