Question

At 517 mm Hg and 24 °C, a sample of gas occuples a volume of 95 ml. The gas is transferred to a 225-ml flask and the temperature is reduced to -
8.0 °C. What is the pressure of the gas in the flask in mmHg?

Answer 1

Answer:

$P_2=194.78mmHg$

Explanation:

Hello,

In this case, we employ the combined ideal gas law in order to understand the volume-gas-pressure behavior as shown below:

$\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

Hence, solving for the final pressure P2, we obtain (do not forget temperature must be absolute):

$P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{517mmHg*95mL*(-8.0+273.15)K}{(24+273.15)K*225mL}\\ \\P_2=194.78mmHg$

Best regards.

Answer 2

Answer:

195mmHg is the pressure of the flask

Explanation:

Combined gas law defines the relationship of pressure, absolute temperature and volume of a gas under different conditions. The formula is:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

In the problem, initial conditions of the gas are:

517mmHg = P₁

24°C + 273.15 = 297.15K = T₁

95mL = V₁

And final conditions are:

225mL = V₂

8.0°C + 273.15 = 265.15K = T₂

Replacing:

$\frac{517mmHg*95mL}{297.15K} =\frac{P_2*225mL}{265.15K}$

P₂ = 195mmHg is the pressure of the flask