**Q1: If 4L of H**

_{2}gas at 1.43 atm is at standard temperature and the pressure was to increase by a factor of 2/3, what is the final volume of the H_{2}gas? (Hint: Boyle's Law)__Solution__

In order to solve this question you need to use Boyle's Law: P

_{1}V_{1}= P_{2}V_{2}Keeping the key variables in mind, temperature and the amount of gas is constant and therefore can be put aside, the only ones necessary are:

- Initial Pressure: 1.43 atm
- Initial Volume: 4 L
- Final Pressure: 1.43x1.67 = 2.39
- Final Volume(unknown): V
_{2}

Plugging these values into the equation you get: V

_{2}= (1.43atm x 4 L)/ (2.39atm) =__2.39 L__**Q2: If 1.25L of gas exists at 35**

^{o}C with a constant pressure of .70 atm in a cylindrical block and the volume were to be multiplied by a factor of 3/5, what is the new temperature of the gas? (Hint: Charles's Law)__Solution__

In order to solve this question you need to use Charles's Law: V

_{1}/T_{1}= V_{2}/T_{2}Once again keep the key variables in mind. The pressure remained constant and since the amount of gas is not mentioned, we assume it remains constant. Otherwise the key variables are:

- Initial Volume: 1.25 L
- Initial Temperature: 35
^{o}C + 273.15 = 308.15K - Final Volume: 1.25L*3/5 = .75 L
- Final Temperature: T
_{2}

Since we need to solve for the final temperature you can rearrange Charles's: T

_{2 }= (T_{1}X V_{2}) / V_{1}Once you plug in the numbers, you get: T

_{2}= (308.15 K x .75 L)/ (1.25 L) =__184.89 K__**Q3: At 655 mm Hg and 25.0**

^{o}C, a sample of Chlorine gas has volume of 750mL. How many moles of Chlorine gas at this condition?__Solution__

P=655mm Hg; T=25+273.15K; V=750mL=0.75L

n = (PV) / (RT)

= [655 mmHg x (1 atm / 760 mmHg)] [0.75 L] / [0.0821 L. atm / mol. K] [25 + 273.15 K]

=

__0.026 mol__

**Q4:**

**5.0 g of neon is at 256 mm Hg and at a temperature of 35º C. What is the volume?**

__Solution__

*Step 1: Write down your given information:*P = 256 mmHg; V =?; m = 5.0 g; R = 0.0821 L·atm·mol

^{-1}K

^{-1}; T = 35º C.

*Step 2: Convert as necessary: 256 mmHg = 0.336*

*8 atm; 5.0 g Neon = 5.0/ 20.18*

*g = 0.2*

*48*

*mol;*

**35º C = 308**

**.15**

**K.**

*Step 3: Plug in the variables into the appropriate equation:*

*V = (nRT) / (P) =*

__18.63__

__L__

*Q**5**:***Find the volume, in mL, when 7.00 g of O2 and 1.50 g of Cl2 are mixed in a container with a pressure of 482 atm and at a temperature of 22º C.**

__Solution__

*Step 1: Write down your given information:*P = 482 atm; V = ?; n = ?; R = 0.0821 L·atm·mol

^{-1}K

^{-1}; T = 22º C + 273.15 = 295.15K; 1.50g Cl

_{2}; 7.00g O

_{2}

*Step 2: Find the total moles of the mixed gases:*n

_{total}= n O

_{2}+ n Cl

_{2}

=

_{ }[7.00 g (1 mol O_{2}/ 32.0 g O_{2})] + [1.50 g (1 mol Cl_{2}/ 70.9 g Cl_{2})]= 0.24 mol

*Step 3: Now that you have moles, plug in your information in the Ideal Gas Equation:*

*V = (nRT) / (P) = 0.012 L =*

__12__

__mL.__

**Q6: A 3.00 L container**

**is filled with Ne (g) at 770mmHg at 27**.

^{o}C. A 0.633g sample of CO_{2}vapor is then added**What is the partial pressure of CO**

_{2}and Ne in atm? What is the total pressure in the container in atm?

__Solution__

*Step 1:*

*Write down all given information*,

*and convert as necessary*: P = 770mmHg --> 1.01atm; V = 3.00L; n(Ne)=?; T = 27

^{o}C --> 300.15K; n(CO

_{2})= ?

n (CO

_{2}) = 0.633 g CO_{2}(1 mol / 44 g CO_{2}) = 0.144 mol CO_{2}*Step 2: Find the unknown moles of Ne.*

*n (Ne) = PV / RT = 0.123 mol*

*Step 3: Now that have pressure for Ne, you must find the partial pressure for CO*.

_{2}. Use the ideal gas equation(PV / nRT)

_{Ne }= (PV / nRT)_{CO2}*Since Volume (V) and Temperature (T) and the Gas Constant (R) are constants, you may remove them from the equation.*

(P / n)

_{Ne }= (P / n)_{CO2}, hence, (1.01 atm / 0.123 mol_{ Ne}) = (P / 0.144 mol_{CO2}), hence, P_{ CO2}= 0.118 atmTherefore, P (total) = P (Ne) + P (CO

_{2}) = 1.01 atm + 0.118 atm =__1.20 atm__**Q7: What is the density of nitrogen gas (N**

_{2}) at 248.0 Torr and 18º C?

__Solution__

*Step 1: Write down your given information:*P = 248.0 Torr --> 0.326 atm; V = ?; n = ?;

R = 0.0821 L·atm·mol

^{-1}K^{-1}; T = 18º C --> 291.15 K1 mol of nitrogen is 14.01 g, hence 2 moles equal to 28.02 g

*Step 2: We need to manipulate the Ideal Gas Equation to incorporate density into the equation.*

d = m/v = P

*M*/RTd = 0382 g/L

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