If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected? a gas at 300 k and 4.0 atm is moved to a new location with a temperature of 250 k. the volume changes from 5.5 l to 2.0 l. what is the pressure of the gas at the new location? what is the partial pressure of 0.50 mol ne gas combined with 1.20 mol kr gas at a final pressure of 730 t

a) Under constant pressure conditions, volume is directly proportional to the temperature. i.e,

V α T

V1/T1 = V2/T2   (1)

It is given that: V2 = 1/3 V1

Substituting for V2 in equation (1) gives:

1/T1 = 1/3T2

i.e. T2 = (1/3) T1

2) Based on the ideal gas equation:

PV = nRT

Under constant n and R

P1V1/T1 = P2V2/T2

P2 = (P1V1/T1)*T2/V2 = (4*5.5/300)*250/2 = 9.17 atm

3) As per Raoult's Law, partial pressure of a gas is equal to the product of mole fraction and the total gas pressure

Mole fraction of Ne = 0.50/0.50 + 1.20 = 0.294

Total pressure = 730 torr

Partial Pressure of Ne = 0.294*730 = 215 torr

If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected?

✔ it would decrease threefold

A gas at 300 K and 4.0 atm is moved to a new location with a temperature of 250 K. The volume changes from 5.5 L to 2.0 L. What is the pressure of the gas at the new location?

✔ 9.2 atm

What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?

✔ 215 torr

Explanation:

there you go boo YW :)