A sample of gas at 47C and 1.5 pressure occupies a volume of 2.20L. What volume would this gas occupy at 107C and 2.5 pressure?

A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of

Rainbow [258]

Answer: (a). T = 38.2 °C (b). V = 1.3392 cm³ (c). ii and iii

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

= yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

= 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

n = PV/RT

n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

4 0

3 years ago

What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?

Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0

3 years ago

Which of the following correctly pairs a phase of

Harrizon [31]

Answer:

njhhhhhhh

Explanation:

8 0

3 years ago

Read 2 more answers

If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure

CaHeK987 [17]

Answer : The volume of gas will be 29.6 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 12 atm

$P_2$ = final pressure of gas = 14 atm

$V_1$ = initial volume of gas = 23 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 200K

$T_2$ = final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{12atm\times 23L}{200K}=\frac{14\times V_2}{300K}$

$V_2=29.6L$

Therefore, the new volume of gas will be 29.6 L

5 0

3 years ago

Read 2 more answers

Please help !!

ipn [44]

Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy = $\frac{22.4L}{1mol}\times 0.545mol=12.208L$ of volume

Hence, 12.208 L of carbon dioxide will be formed.

5 0

3 years ago