What is the volume of an original copper rod

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

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

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

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

$\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}$

$P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}$

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0

2 years ago

A fuel tank holds 22.3 gallons of gasoline. If the density is 0.8206 g/mL, what is the mass in kilograms of gasoline in a full t

7nadin3 [17]

Answer:

$m=69.3kg$

Explanation:

Hello!

In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

$V=22.3gal*\frac{3.78541L}{1gal}=84.415L$

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

$m=0.8206g/L*84.415L\\\\m=69.3kg$

Best regards!

8 0

3 years ago