1.A spray can is used untilit s empty except for the propellant gas, which has a

Question

1.A spray can is used untilit s empty except for the propellant gas, which has a pressure of 1344 torr at 23°C. If the can is thrown into a fire(T=475 °C), what will be the pressure in the hot can? (3 pts) 2. What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 23.5 L at 65 C and 744 torr? (3 pts) 3. A 2.50-L volume of hydrogen measured at-100 °C is warmed to 100 °C Calculate the volume of the gas at the higher temperature, assuming no change in pressure.(3 pts) 4. Define the ideal gas law (2 pts) 5. How many grams of gas are present in each of the following cases? (6 pts) 2.0.500 L of CO2 at 507 torr and 36 °C b. 221 mL of He at 0.23 torr and 54 °C 6. If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure? (2 pts) 7. If the volume of a fixed amount of a gas is doubled at constant temperature, what happens to the pressure? (2 pts)

Explanation / Answer

2.

Charles’s law states that when the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be directly related.

V1/T1 = V2/T2

Here pressure = 744 torr is maintained constant

V1 = 11.2 L

T1 = ?

V2 = 23.5 L

T2 = 65 oC = (273 + 65)K = 338 K

Now,

V1/T1 = V2/T2

T1 = V1T2/V2

   = (11.2 L) (338 K) / (23.5 L)

    = 161 K

    = (161 – 273) oC

    = - 112 oC

3.

V1 = 2.5 L

T1 = - 100 oC = (-100 + 273) K = 173 K

V2 = ?

T2 = 100 oC = (273 + 100)K = 373 K

Now,

V1/T1 = V2/T2

V2 = V1T2/T1

   = (2.50 L) (373 K) / (173 K)

    = 5.4 L

4.

Ideal gas law states that the product of the pressure (P) and the volume (V) of an ideal gas is equal to the product of the number of moles of the gas (n), absolute temperature (T) of the gas and the universal gas constant (R).

Mathematically,

PV = nRT

5.

(a)

V = 0.500 L

P = 507 torr

T = 36 oC = (36 + 273)K = 309 K

R = 62.364 L Torr mol-1K-1

Now,

PV = nRT

n = PV / RT

   = [ (507 torr) (0.500 L) ] / [ (62.364 L Torr mol-1K-1) (309 K) ]

   = 0.013 mol

Molar mass of CO2 = 44 g/mol

So, 1 mol of CO2 = 44 g

0.013 mol of CO2 = 0.013 x 44 g = 0.572 g

(b)

V = 221 mL = 0.221 L

P = 0.23 torr

T = 54 oC = (54 + 273) K = 327 K

R = 62.364 L Torr mol-1K-1

Now,

PV = nRT

n = PV / RT

   = [ (0.23 torr) (0.221 L) ] / [ (62.364 L Torr mol-1K-1) (327 K) ]

   = 2.5 x 10-6 mol

Molar mass of He = 4 g/mol

So, 1 mol of He = 4 g

(2.5 x 10-6) mol of He = (2.5 x 10-6) x 4 g = (1.0 x 10-5) g

6.

PV = nRT

Here, V, n and R are constant. So, if the temperature is doubled then pressure will also increase by 2 times.

7.

PV = nRT

Here, T, n and R are constant.

So, PV = constant

Hence, if the volume is doubled then pressure will reduced to half (1/2).

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