Question

Calculate the work done during combustion of 0.138 kg of ethanol, $C_{2}H_{5}OH_{\left(\right. l \left.\right)}$ at 300 K. (Given: $R=8.314JK^{- 1}mol^{- 1}$ , molar mass of ethanol $=46gmol^{- 1}$ .)

• A. - 7482 J
• B. 7482 J
• C. - 2494 J
• D. 2494 J

Answer 1

Answer: (B)

$C_{2}H_{5}OH_{\left(\right. l \left.\right)}+3O_{2 \left(\right. g \left.\right)} \rightarrow 2CO_{2 \left(\right. g \left.\right)}+3H_{2}O_{\left(\right. l \left.\right)}$ …(i)

0.138 kg of ethanol $\equiv \frac{138 g}{46 g m o l^{- 1}}=3$ moles

By multiplying equation (i) by 3, we get

$3C_{2}H_{5}OH_{\left(\right. l \left.\right)}+9O_{2 \left(\right. g \left.\right)} \rightarrow 6CO_{2 \left(\right. g \left.\right)}+9H_{2}O_{\left(\right. l \left.\right)}$

$\Delta n_{g}=6-9=3$

$w=-P\Delta V$

$w=-\Delta n_{g}RT=-\left(\right.-3\left.\right)\times 8.314\times 300$ $\left(\right.\because PV=nRT\left.\right)$

$=7482.6$ J