The temperature of a certain mass of gas was increased from $40^{o} C$ to $41^{o} C$ at constant pressure. The volume of the gas:

will remain constant

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will increase by

\frac{1}{273}

of its volume at

273 K

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will increase by

\frac{1}{313}

of its volume at

40^{o} C

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will increase, but the increase in volume cannot be predicted

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Solution

The correct option is C will increase by $\frac{1}{313}$ of its volume at $40^{o} C$
Let $V_{i}$ and $V_{f}$ represents initial and final volumes respectively.
$T_{i}$ and $T_{f}$ will represent initial and final temperatures respectively.
$T_{i} = 40^{o} C = 40 + 273 = 313 K$
$T_{2} = 41^{o} C = 41 + 273 = 314 K$
The ideal gas equation is $P V = n R T$
Since, the pressure and the number of moles of gas are constant, $\frac{V_{i}}{T_{i}} = \frac{V_{f}}{T_{f}}$
Substitute values in the above expression.

$\frac{V_{i}}{313 K} = \frac{V_{f}}{314 K}$

Hence, $V_{f} = \frac{314 K}{313 K} \times V_{i} = 1 + \frac{1}{313} \times V_{i}$

Thus, the volume of the gas will increase by $\frac{1}{313}$ of its volume at $40^{o} C$

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Similar questions

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The temperature of a certain mass of gas was increased from 40oC to 41oC at constant pressure. The volume of the gas:

The temperature of a certain mass of gas was increased from $40^{o} C$ to $41^{o} C$ at constant pressure. The volume of the gas: