A 2-00 mole sample of an ideal gas is placed into a rigid 2-00L vessel 1-00 atm-What will the temperature of the gas be when the volume of the gas is reduced to 0 L-273-circ-C-273-circ-C100-circ-C0-circ-C

Question

Toppr Admin · Accepted Answer

N-2V1-2lP1-1atmT2T1-P2V2P1V1asV2-0lT2-0KT1-x2212-273-x2103-Ans -x2212-273-x2103-

A $2.00$ mole sample of an ideal gas is placed into a rigid $2.00 L$ vessel at $1.00 a t m$ .
What will the temperature of the gas be when the volume of the gas is reduced to $0 L$ ?
$273^{\circ} C$
$- 273^{\circ} C$
$100^{\circ} C$
$0^{\circ} C$

$n = 2 V_{1} = 2 l P_{1} = 1 a t m \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}} a s V_{2} = 0 l T_{2} = 0 K T_{1} = - 273 ℃$
Ans $= - 273 ℃$

A 2.00 mole sample of an ideal gas is placed into a rigid 2.00L vessel at 1.00atm.What will the temperature of the gas be when the volume of the gas is reduced to 0 L?273∘C−273∘C100∘C0∘C

n=2V1=2lP1=1atmT2T1=P2V2P1V1asV2=0lT2=0KT1=−273℃Ans =−273℃

A $2.00$ mole sample of an ideal gas is placed into a rigid $2.00 L$ vessel at $1.00 a t m$ .
What will the temperature of the gas be when the volume of the gas is reduced to $0 L$ ?
$273^{\circ} C$
$- 273^{\circ} C$
$100^{\circ} C$
$0^{\circ} C$

$n = 2 V_{1} = 2 l P_{1} = 1 a t m \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}} a s V_{2} = 0 l T_{2} = 0 K T_{1} = - 273 ℃$
Ans $= - 273 ℃$