Answer to Question #164325 in General Chemistry for muna mohamed

Question

1. A gas is at 135°C and 455 mm Hg in a 2.00 L container. It is cooled down to a temperature of 25°C. If it is kept in the same container, what is its new pressure?
2. A gas has a volume of 39 L at STP. What will its volume be at 4 atm and 25°C?
3. A gas, now contained at STP, has a volume of 500 mL. The initial pressure was 0.96 atm at 20°C. What was the initial volume?
4. A balloon is filled up with air to a volume of 1.15 L at 296.5 K. What does the volume change to if the balloon is taken outdoors where the temperature is 278.4 K and the pressure is half of what it was indoors?
5. Calculate the pressure in a tire if it starts out filled with 7.54 L of air at 219 kPa and 21.6°C and gets heated to 65.2°C as the volume increases to 7890 mL.
6. A 700.0 mL gas sample gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.00°C. What is the new pressure of the gas in kPa?
7. A balloon of air now occupies 10.0 L at 25.0°C and 1.00 atm. What temperature was it initially, if it occupied 9.4 L and was in a freezer with a pressure of 0.939 atm?
8. 6.8 L of a gas is found to exert 97.3 kPa at 25.2°C. If the volume is changed to 12.5 L, what would be the required temperature to change the pressure to standard pressure?
9. A cylinder is fitted with a moveable piston. A gas in the cylinder is heated up. If the pressure only increases slightly, what will happen to the volume of the cylinder? Why?

Answer 1

Q164325

1. Gas is at 135°C and 455 mm Hg in a 2.00 L container. It is cooled down to a temperature of 25°C. If it is kept in the same container, what is its new pressure?

Solution:

The gas is present in the same container, so the volume will remain constant.

Amontons law relates Pressure and temperature.

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

P₁ = 455 mm Hg

T₁ = 135°C = 135+273.15 = 408.15 K

P₂ = unknown.

T₂ = 25°C = 25+273.15 = 298.15K

plug all this information in the formula we have

$\frac{455 mm Hg }{P_2}=\frac{408.15K}{298.15K }$

Arranging this equation for P₂ we have

$P_2 = \frac{408mmHg* 298.15K}{408.15K} = 332.4 mm Hg$

Hence the new Pressure of the gas will be 332 mm Hg.

2) A gas has a volume of 39 L at STP. What will its volume be at 4 atm and 25°C?

Solution:

At STP, P = 1atm and T = 273.15 K

The gas has a volume of 39L at STP.

Sp P₁ = 1atm. V₁ = 39L, T₁ = 273.15K

We have to find the volume occupied by the gas at 4 atm and 25°C

V₂ = unknown, P₂ = 4 atm . T₂ = 25°C = 25 + 273.15 = 298.15K

We will use the combined gas equation here.

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

$\frac{1atm*39L}{273.15K}=\frac{4atm*V_2}{298.15K}$

arranging this equation for V2 we have

$V_2= \frac{1atm*39L*298.15K}{273.15K*4atm }= 10.64L$

which in 2 significant figures is 11 L.

Hence the volume occupied by the gas at 4 atm and 25°C is 11 L.

Question3

A gas, now contained at STP, has a volume of 500 mL. The initial pressure was 0.96 atm at 20°C. What was the initial volume?

Solution:

This problem is similar to the second problem.

At STP, P = 1atm and T = 273.15 K

The gas has a volume of 500 mL at STP.

So P_f = 1atm. V_f = 500 mL, T_f = 273.15K

We have to find the volume occupied by the gas at 0.96 atm and 20°C

V_i = unknown, P_i = 0.96 atm . T_i = 20°C = 20 + 273.15 = 293.15K

We will use the combined gas equation here.

$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$

$\frac{0.96atm*V_i}{293.15K}=\frac{1atm*500mL}{273.15K}$

arranging this equation for Vi we have

$V_i=\frac{1atm*500mL*293.15K}{273.15K*0.96atm} = 559 mL$

which in 2 significant figures is 560 mL.

Hence the initial volume of the gas at initial pressure 0.96 atm and initial temperature 20°C

was 560mL.

Question 4.

A balloon is filled up with air to a volume of 1.15 L at 296.5 K. What does the volume change to if the balloon is taken outdoors where the temperature is 278.4 K and the pressure is half of what it was indoors?

Solution:

We will again use the same combine equation of gas.

outdoors pressure is half of the indoor pressure.

This means P1 = 2 P2 or P1 / P2 = 2/1

We will use the combined gas equation here.

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

arranging this equation for V₂ we have

$V_2=\frac{P_1V_1T_2}{T_1P_2}$

$V_2= \frac{2*1.15L*278.4K}{296.5K }= 10.64L =2.160L$

which in 3 significant figures is 2.16 L.

Hence the balloon will occupy 2.16L when it is taken outdoor.

5) Calculate the pressure in a tire if it starts out filled with 7.54 L of air at 219 kPa and 21.6°C and gets heated to 65.2°C as the volume increases to 7890 mL.

Solution :

P₁ = 219 kPa, V₁ = 7.54 L, T₁ = 21.6°C = 21.6 + 273.15 = 294.75K

P₂ = unknown,

V₂ = 7890 mL = 7890mL * 1L/ 1000 mL = 7.890 L,

T₂ = 65.2°C = 62.2 + 273.15 = 338.35K

We will use the combined gas equation here.

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

arranging this equation for P₂ we have

$P_2=\frac{P_1V_1T_2}{T_1V_2}$

$P_2= \frac{219 kPa*7.54L*338.35K}{294.75K*7.890L }= 240.24kPa =240 kPa.$

Hence the pressure in the tire will be 240 kPa.

6) A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.00°C. What is the new pressure of the gas in kPa?

Solution:

700.0mL gas sample is present at STP.

Sp V₁ = 700.0mL, T₁ = 273.15K, P₁ = 1 atm.

the new volume after compression is

V₂ = 200.0mL and new temperature is T₂ = 30.00 °C

T₂ = 30.00 °C + 273.15 = 303.15K

We have to find the new pressure.

We will use the combined gas equation here.

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

arranging this equation for P₂ we have

$P_2=\frac{P_1V_1T_2}{T_1V_2}$

$P_2= \frac{1atm*700.0mL*303.15K}{273.15K*200.0mL }= 3.88 atm.$

which in kPa is

$New\space pressure\space of\space gas\space in\space kPa = 3.88 atm * \frac{101.325kPa}{1atm} = 394kPa.$

so the new pressure of the gas will be 394 kPa.

7) A balloon of air now occupies 10.0 L at 25.0°C and 1.00 atm. What temperature was it initially, if it occupied 9.4 L and was in a freezer with a pressure of 0.939 atm?

Solution :

Vf = 10.0L, Tf = 25.0°C + 273.15 = 298.15K , Pf = 1.00 atm.

Vi = 9.4L , Ti = unknown, Pi = 0.939 atm

PiVi/Ti = PfVf/ Tf

arranging this equation for Ti we have

Ti = PiViTf / Pf Vf )

Ti = ( 0.939atm * 9.4L * 298.15K)/(1.00atm * 10.0L)

T i = 263.17 K

which in degree celsius is 263.17 - 273.15 = -9.985 °C = -10.0 °C

Hence the initial temperature of the air inside the balloon was -10.0 °C

8) 6.8 L of a gas is found to exert 97.3 kPa at 25.2°C. If the volume is changed to 12.5 L, what would be the required temperature to change the pressure to standard pressure?

Solution:

P1 = 97.3 kPa, V1 = 6.8L, T1 = 25.2°C = 25.2 + 273.15 = 298.35K

P2 = stand pressure = 1atm = 101.325 kPa . V2 = 12.5 L

and T2 = unknown.

P1V1/T1 = P2V2/T2

arranging this equation for T2 we have

T2 = P2V2T1/ P1V1

T2 = (101.325kPa* 12.5L* 298.35K)/(97.3kPa* 6.8L)

T2 = 571.12 Kelvin

which in degree celsius is 571.12 - 273.15 = 297.8 °C

Hence the temperature required will be 297.8 °C

9) A cylinder is fitted with a moveable piston. A gas in the cylinder is heated up. If the pressure only increases slightly, what will happen to the volume of the cylinder? Why?

Solution:

We are increasing the temperature and the pressure is increased slightly.

Volume is directly proportional to temperature and inversely proportional to pressure.

In case if pressure is only increased slightly then the volume will increase with the increase in temperature.

This is because the volume is directly proportional to temperature

V ∝ T.