Question

To what temperature does a 250.0-ml cylinder containing 0.1000 mol he need to be cooled in order for the pressure to be 253.25 kpa

Answer 1

76.19 K according to Gradpoint (my hw).

Answer 2

T = 76.09 K

Step-by-step explanation:

Given:

Moles of He, n = 0.1000

Volume, V = 250.0 ml = 0.250 L

Pressure, P = 253.25 kPa

To determine:

Temperature, T

Formula:

Based on the ideal gas law: PV = nRT -----(1)

R = 0.0821 L.atm/mol-K

P = 253.25 kPa

Now, 1 kPa = 0.00987 atm

Therefore, 253.25 kPa = 2.499 atm

Based on equation (1)

`T = (PV)/(nR) = (2.499 atm*0.250 L)/(0.1000 mol *0.0821 Latm/mol-K)=76.09 K`