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# Half mole of an ideal mono-atomic gas is heated at constant pressure of $1\,atm$ from $20^\circ C$ to $90^\circ C$. Work done by gas is close to: (Gas constant $R = 8.31J/mol.K$ )A) $73J$B) $291J$C) $581J$D) $146J$

Last updated date: 16th Jun 2024
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Hint: We are given that the gas used is ideal gas. Use the principles and concepts of ideal gas in the solution and transform the usual formula of work done by a gas such that it coincides with the values we are provided with.

Formula Used:
Work Done by a gas, $WD = P\Delta V$
Where, $P$ is the pressure during the reaction and $\Delta V$ is the change in volume of the gas during the reaction
For an ideal gas, $P\Delta V = nR\Delta T$
Where, $P$ is the pressure during the reaction, $\Delta V$ is the change in volume of the gas during the reaction, $n$ is the number of moles of the gas, $R$ is the Gas Constant (value given in question), $\Delta T$ is the change in temperature of the gas during the reaction

Complete Step by Step Solution:
We are given that during the reaction half mole of the gas is used, the pressure is $1\,atm$ , and the temperature changes from $20^\circ C$ to $90^\circ C$. Now, we know that work done by a gas, $WD = P\Delta V$.
Also, the gas used is ideal. We know that in the case of an ideal gas, $P\Delta V = nR\Delta T$
Hence, we conclude that $WD = nR\Delta T$
On putting the respective values, we get $WD = 0.5 \times 8.31 \times (90 - 20)$
Simplifying, $WD = 0.5 \times 8.31 \times 70$
$WD = 0.5 \times 581.7 = 290.85 \simeq 291J$

Hence, option B is the correct answer.

Note: Do not forget to check the number of moles in the question. Like in this question, we are given that half mole of the gas has been used. We often ignore this and end up solving the question with one mole which makes our answer incorrect. Questions like these mostly have an option with the answer that you would have got if the number of moles was one (like in this question option C). Pay attention to the given values.