Rodesaespiritu6idn
Answered

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1 .If I have 4 moles ofa gas at a pressure of 5.6 atm and a volume of 12 liters ,what
is the temperature ?
2. If have an unknown quantity gas at pressure of 1.2 atm ,a volume 31
liters ,and a temperature of 87o C ,how many moles gas do I have ?
3. If I contain 3 moles ofgas in a container with a volume of liters and at a
temperature of 400 K ,what is the pressure inside the container ?
4. If I have 7.7 moles of gas at a pressure of 0.09 atm and at atemperature of 56 C ,
what is the volume of the container that the gas is in ?
5 .If have 17 moles of gas at a temperature of 67 C ,and a volume of 88.89 liters
what is the pressure ofthe gas ?
6. If I have an quantity of gas at a pressure of 0.5 atm .a volume of 25
liters ,and a temperature of 300 K ,how many moles of gas do I have ?)
7 .If I have moles of gas held at a pressure of 78 atm and a temperature of K
what is the volume of the gas ?
8. If I have 1.9 moles of gas held at a pressure of 5 atm and in a container with a
volume of 50 liters ,what is the temperature of the gas ?
9. If I have 2.4 moles of gas held at a temperature of C and in a container with a
volume of 45 liters ,what is the pressure of the gas ?
10. If I have an quantity of gas held at a temperature of 1195 K in a
container with a volume of 25 liters and a pressure of 560 ,how many moles
of gas do I have ?
​

Sagot :

Rosedeguzman224idn

1. The temperature is 136K.

S0LÙTÍ0N :)

PV=n

RTn=6mol

P=5,6atm

V=12lR=0.0821

T=?

T=P V/nR= (5.6*12)/(6*0.0821)=67.2/0.4926=136K

2. The moles gas do you have is 2.52.

S0LÙTÍ0N :)

PV=n

RTP=1.8atm

V=41l

T=87°=357

KR=0.0821

n=?

n=PV/RT=(1.8*41)/(0.0821*357)=73.8/29.3=2.52mol

3. 1.642 atmosphere

S0LÙTÍ0N :)

n = 3.0 moles

V = 60.0

LT = 400 K

From PV = n

RT, you can find PP = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L

4. 2310.89 L or 2.3 KL

S0LÙTÍ0N :)

n = 7.7 moles

n = 7.7 molesP = 0.090 atm

n = 7.7 molesP = 0.090 atmT = 56ºC = 329ºK

5. The new pressure will be 140 atm

S0LÙTÍ0N :)

[tex] \large \: p2 = \large \frac{p1v1t2}{v2t1 } [/tex]

[tex]p2 = \frac{88.89atom \times 17l \times 367k}{12l \times 340k} \\ = 140 \: atom[/tex]

6. Thus, the moles of the gas are 0.508mol.

S0LÙTÍ0N :)

[tex] \large \: n = \frac{(0.5 \: atm)(25.0 \: l)}{(0.0821l. \: atm (mol.k))(300k) } [/tex]

[tex] \large \: = 0.5075111652 \: mol[/tex]

[tex] \large \: ≃ \: 0.508 \: mol[/tex]

[tex]\huge\color{green}carry \: on \: learning[/tex]