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IRISSAK [1]
2 years ago
5

In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the

mixture is 75.0 torr, what is the total pressure of the mixture
Chemistry
1 answer:
notka56 [123]2 years ago
4 0

Answer:

P_T=112.4torr

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

P_{H_2}=x_{H_2}*P_T

And can be solved for the total pressure as follows:

P_T=\frac{P_{H_2}}{x_{H_2}}

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667

Then, we can plug in to obtain the total pressure:

P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr

Regards!

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A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr
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Answer:

184.62 ml

Explanation:

Let p_1, v_1, and T_1 be the initial and p_2, v_2, and T_2 be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

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T_1= 26 ^{\circ}C = 273+26 =299 K

T_2= 3 ^{\circ}C = 273+3 =276 K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)

For the final condition,

p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)

Equating equation (i), and (ii)

\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

\frac{v_1}{T_1}=\frac{v_2}{T_2}  [from equation (i)]

v_2=\frac{T_2}{T_1} \times v_1

Putting all the given values, we have

v_2=\frac{276}{299} \times 200 = 184.62 \; ml

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

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