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]
