Question

What pressure (in atm) would be exerted by a mixture of 1.4 g of nitrogen gas and 4.8 g of oxygen
gas in a 200 mL container at 57°C?
(a) 4.7
(b) 34
(c) 47
(d) 27​

Answer 1

Answer:

(d) 27 atm

Explanation:

Answer 2

Answer:

The pressure, P, exerted by the mixture of gases measured is $27atm$.

Therefore, option d) $27atm$ is correct.

Explanation:

Given,

The mass of nitrogen gas ( $N_{2}$ ) = $1.4g$

The mass of oxygen gas ( $O_{2}$ ) = $4.8g$

The volume of gas, V = $200ml$ = $0.2L$

The temperature of the gas, T = $57\textdegree C$ = $(57 + 273)K$ = $330K$

The pressure exerted by the mixture of gases, P =?

As we know,

• From the ideal gas equation;
• $PV =n RT$      
• $P = \frac{nRT}{V}$                 -------equation (1)

Here,

• n = The number of moles
• R = The gas constant = $0.0821atm-L-K/mol$

Now, we have to calculate the total number of moles.

As we know,

• The molar mass of $N_{2}$ = $28g/mol$
• The molar mass of $O_{2}$ = $32g/mol$

Now,

• The number of moles of $N_{2}$ = $\frac{Given mass}{Molar mass}$ = $\frac{1.4}{28}$ = $0.05mol$
• The number of moles of $O_{2}$ = $\frac{Given mass}{Molar mass}$ = $\frac{4.8}{32}$ = $0.15mol$

Therefore, the total number of moles in the mixture of gases = $0.05+ 0.15$ =          $0.2mol$.

After putting all the values in the equation (1), we get:

• P = $\frac{0.2\times 0.0821\times 330}{0.2}$
• P = $27atm$

Hence, the pressure exerted by the gas, P = $27atm$.