# Iam stocked please guys help me out with this chemistry

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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?

10. You go on a road trip to escape the cold winter days of Kirkland Lake, Ontario to the warm beaches of Moonbeam, Ontario. Since it is a long drive and you are leaving early in the morning, you make sure to check the tire pressure and make sure it is at the recommended level.
At lunchtime, you stop for gas and decide to check your car’s tire pressure again. You also notice that the tires seem to be a little bit larger than when you left.
The driver’s manual suggests a pressure between 205 and 220 kPa. Knowing that the tires heat up while driving; do you expect your reading to be high, low or the same as the recommended tire pressure? Why? Since you plan on staying in Moonbeam a while to see the sights, you think it would be best to adjust your tire pressure. What adjustments should be made to your tires? Should you make these adjustments as soon as you arrive, or the following day?

11. Think of a situation where pressure is constant but temperature changes. Discuss why this may seem to violate one of the gas laws. Explain why it does not actually violate the gas laws.

#372908

Dont worry my dear, this are the solutions you are covered

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?

The gas is present in the same container, so the volume will remain constant.
Amontons law relates Pressure and temperature.
P1/P2 = T1/T2
P1 = 455 mm Hg
T1 = 135°C = 135+273.15 = 408.15 K
P2 = ?
T2 = 25°C = 25+273.15 = 298.15K
Plug all this information in the formula we have
(455 mmHg)/P_2 =408.15K/298.15K
Arranging this equation for P2 we have
P2=(408mmHg* 298.15K)/408.15K = 332.4 mm Hg

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

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.
(P1* V1)/T1 = (P2* V2)/T2
(1atm*39L)/273.15=〖4atm * V2]/298.15K
Arranging this equation for V_2 we have
V2= (1atm*39L*298.15)/(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.

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?
At STP, P = 1atm and T = 273.15 K
The gas has a volume of 500 mL at STP.
So Pf = 1atm. Vf = 500 mL, Tf = 273.15K
We have to find the volume occupied by the gas at 0.96 atm and 20°C
Vi = unknown, Pi = 0.96 atm . Ti = 20°C = 20 + 273.15 = 293.15K

We will use the combined gas equation here.
(Pi* Vi)/Ti = (Pf* Vf)/Tf
(0.96 atm*V_i)/293.15K=(1atm*500ml)/273.15K
Arranging this equation for Vi we have
Vi= (1 atm*500mL*293.15K)/(273.15K*0.96 atm)
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.

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?

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.
(P1* V1)/T1 = (P2* V2)/T2
Arranging this equation for V_2 we have
V2= (2*1.15*278.4K)/296.5K = 2.160L
This in 3 significant figures is 2.16 L.

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.

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.
(P1* V1)/T1 = (P2* V2)/T2
P2= (P1* V1* T2)/(T1* V2 )
Arranging this equation for P2 we have
P2= (219kPa * 7.54L *338.35k)/(294.75k *7.890L) = 240.24kpa = 240kpa
Hence the pressure in the tire will be 240 kPa.

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?

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.
(P1* V1)/T1 = (P2* V2)/T2
P2= (P1* V1* T2)/(T1 * V2 )
Arranging this equation for P2 we have
P2= (1atm * 700mL *303.15k)/(273.15k *200.0mL) = 3.88atm
Which in kPa is
New pressure of gas in kPa = 3.88atm * 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?
Vf = 10.0L, Tf = 25.0°C + 273.15 = 298.15K , Pf = 1.00 atm.
Vi = 9.4L , Ti = unknown, Pi = 0.939 atm
We will use the combined gas equation here.
(Pi* Vi)/Ti = (Pf* Vf)/Tf
Arranging this equation for Vi we have
Ti= (0.939 atm *298.15 L * 293.15K)/(1.00 atm * 10.0L)
Ti=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?
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.
(P1* V1)/T1 = (P2* V2)/T2
T2= (P2* V2* 1)/(P1* V1 )
Arranging this equation for T2 we have
T_2= (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?
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.

10) You go on a road trip to escape the cold winter days of Kirkland Lake, Ontario to the warm beaches of Moonbeam, Ontario. Since it is a long drive and you are leaving early in the morning, you make sure to check the tire pressure and make sure it is at the recommended level.
At lunchtime, you stop for gas and decide to check your car’s tire pressure again. You also notice that the tires seem to be a little bit larger than when you left.
The driver’s manual suggests a pressure between 205 and 220 kPa. Knowing that the tires heat up while driving; do you expect your reading to be high, low or the same as the recommended tire pressure? Why? Since you plan on staying in Moonbeam a while to see the sights, you think it would be best to adjust your tire pressure. What adjustments should be made to your tires? Should you make these adjustments as soon as you arrive, or the following day?

Here the second part, the pressure will be high. This is due to the increase in temperature of the tyres caused by the frictional force with the road. When T increases, P increases too. Assuming the others are kept constant. In this case, the tyres are fully closed, suggesting that there are no leakage and the number of moles, n, will be constant. Since the tyre has a rigid shape, the volume of tyre is also constant.
For the next part, i am not every sure but i believe the correct answer should be yes, the pressure should be adjusted. In fact, it should be lowered as soon as possible to prevent rupture of the tyres due to high pressure.

11) Think of a situation where pressure is constant but temperature changes. Discuss why this may seem to violate one of the gas laws. Explain why it does not actually violate the gas laws.