A sample of gas at $35^{o} C$ & 1 $a t m$ pressure occupies a volume of 3.75 litres. At what temperature should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure?
(in $K$ to the nearest integer)

Open in App

Solution

As we know,
$V \propto T$

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

$T_{2} = \frac{3}{3.75} \times 308 \Rightarrow 246.4 K$

$T_{2} = - {26.6}^{o} C$

$T = 273 - 26.6 = 246.4 \equiv 246 K$

Suggest Corrections

Similar questions

Q. A sample of gas at

35^{o} C

& 1 atm pressure occupies a volume of 3.75litres. At what temp. should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure:

Q. If the volume occupied by a gas at 337K and 10 atm is 1 litre, it'll occupy litres, if the pressure is reduced to 1 atm, temperature being same.

Q. A sample of gas at 1 atm occupies volume amounting to 1 litre. Now it expands at constant temperature to 4 litres, find heat added to gas. Take (

ln 2 = 0.693

Q. A gas occupies

V

litres at a certain temperature. Keeping its pressure constant, its volume can be reduced to half its initial volume at

27^{o} C

. The original temperature of the gas is:

At 27 $^{0}$ C certain gas occupied a volume of 4 litres. If the volume of this is to be increased to 12 litres at the same pressure, the gas is to be heated to a temperature

Join BYJU'S Learning Program

A sample of gas at 35oC & 1 atm pressure occupies a volume of 3.75 litres. At what temperature should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure?(in K to the nearest integer)

As we know,V∝TV2V1=T2T2T2=33.75×308⇒246.4KT2=−26.6oCT=273−26.6=246.4≡246 K

A sample of gas at $35^{o} C$ & 1 $a t m$ pressure occupies a volume of 3.75 litres. At what temperature should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure?
(in $K$ to the nearest integer)

As we know,
$V \propto T$

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

$T_{2} = \frac{3}{3.75} \times 308 \Rightarrow 246.4 K$

$T_{2} = - {26.6}^{o} C$

$T = 273 - 26.6 = 246.4 \equiv 246 K$