Question

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.350 atm . Part A What is the partial pressure of argon, PAr, in the flask?

Answer 1

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 $m^{3}$

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 $\frac{J}{kg k}$

From ideal gas equation P V = m RT

⇒ P = $\frac{m R T}{V}$

Put all the values in above formula we get

⇒ P = $\frac{0.00115}{0.001}$ × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa