Two closed vessels of equal volume contain air at $105 k$ $P a$ , $300 K$ and are connected throw a narrow tube. If one of the vessels is now maintained at $300 K$ and the other at $400 K$ , what will be the pressure in the vessels (in $k P a$ )?

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Solution

Let the other pressure be $P$
According to the ideal gas equation
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Volume is same as per the question
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\frac{105}{300} = \frac{P_{2}}{400}$
$P = 140 k P a$
Pressure in the vessel is 140 kPa.

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Two closed vessels of equal volume contain air at 105k Pa, 300K and are connected throw a narrow tube. If one of the vessels is now maintained at 300K and the other at 400K, what will be the pressure in the vessels (in kPa)?

Let the other pressure be PAccording to the ideal gas equationP1V1T1=P2V2T2Volume is same as per the questionP1T1=P2T2105300=P2400P=140kPaPressure in the vessel is 140 kPa.

Two closed vessels of equal volume contain air at $105 k$ $P a$ , $300 K$ and are connected throw a narrow tube. If one of the vessels is now maintained at $300 K$ and the other at $400 K$ , what will be the pressure in the vessels (in $k P a$ )?

Let the other pressure be $P$
According to the ideal gas equation
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Volume is same as per the question
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\frac{105}{300} = \frac{P_{2}}{400}$
$P = 140 k P a$
Pressure in the vessel is 140 kPa.