A volume 'V' and temperature 'T' was obtained, as shown in the diagram, when a given mass of gas was heated. During the heating process, the pressure is

increased

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decreased

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remains constant

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changed erratically

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Solution

The correct option is B decreased
$\frac{d V}{d T} = c o n s t a n t$ from the plot.
But, $\frac{d V}{d T} = T + \frac{d P}{d T}$
$∴ \frac{d P}{d T} = C o n s t a n t - T$
$∴$ P decreases as T increases.

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A volume 'V' and temperature 'T' was obtained, as shown in the diagram, when a given mass of gas was heated. During the heating process, the pressure is

The correct option is B decreaseddVdT=constant from the plot.But, dVdT=T+dPdT∴dPdT=Constant−T∴ P decreases as T increases.

The correct option is B decreased
$\frac{d V}{d T} = c o n s t a n t$ from the plot.
But, $\frac{d V}{d T} = T + \frac{d P}{d T}$
$∴ \frac{d P}{d T} = C o n s t a n t - T$
$∴$ P decreases as T increases.