Arigid canister holds 50.0 l of helium gas at 32.0°c and 49.9 kpa. the gas is cooled until the pressure reaches 14.2 kpa. what is the kelvin temperature of the helium?

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86.794 K

Explanation:-

Initial temperature T 1 = 32 C + 273 = 305 K

Initial Pressure P 1 = 49.9 KPa

New Pressure P 2 = 14.2 KPa

The Volume remains same since it is rigid.

∴V 1 = V 2 = 50.0 L

Using the relation

P 1 V 1 / T1 = P 2 V 2 / T 2

New temperature T 2 = P 2 V 2 T 1 / ( P 1 V 1)

= 14.2 KPa x 50.0 L x 305 K / ( 49.9 KPa x 50.0 L)

= 86.794 K

po-215 i got it right on the quiz