A 7.70 L container holds a mixture of two gases at 51 ° C. The partial pressures of gas A and gas B, respectively, are 0.177 atm and 0.726 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Respuesta :

Answer:

[tex]P_{total}=1.628atm[/tex]

Explanation:

According to dalton theory of partial pressure , partial pressure applied by a component in a mixture of more gas is same as pressure applied when it is taken alone in the container.

And thus [tex]P_{total}=P_A+P_B+P_C[/tex]

and P_A and P_B is given so we need to find P_C for the 3rd gas,

and we will solve by assuming that gas follows the ideality nature of gas,

PV=nRT

P_A=0.177atm

P_B=0.726atm

[tex]P_C \times 7.7L =0.21 \times 0.0821 \times 324[/tex]

[tex]P_C=0.725atm[/tex]

therefore,

[tex]P_{total}=P_A+P_B+P_C[/tex]

[tex]P_{total}=0.177atm +0.726atm + 0.725atm[/tex]

[tex]P_{total}=1.628atm[/tex]

Answer:

0.12

Explanation:

Di the test