Question

Work done by 0.1 mole of a gas at 27 degree celsius to double its volume at constant pressure is...? (r= 2 cal/mol)​

Answer 1

Answer:

Explanation: i have no clue

Answer 2

The work done is 60 J.

Explanation:

Given that,

Temperature = 27°C

Volume $V_{2}= 2V_{1}$

Number of mole = 0.1

We need to calculate the work done

Using equation of ideal gas

$PV=nRT$

$V=\dfrac{nRT}{P}$

Here, Pressure is constant

$V\propto T$

$\dfrac{V_{1}}{V_{2}}=\dfrac{T_{1}}{T_{2}}$

Put the value of V₂

$\dfrac{V_{1}}{2V_{1}}=\dfrac{T_{1}}{T_{2}}$

$T_{1}=2T_{2}$

Now using formula of work done

$W=P\ cdot \Delta V$

$W=nR\Delta T$

Put the value into the formula

$W=nR(T_{2}-T_{1})$

$W=nR(2T_{1}-T_{1})$

$W=nRT_{1}$

Put the value into the formula

$W=0.1\times2\times300$

$W=60\ J$

Hence, The work done is 60 J.

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Topic : work done

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