Answer:
Water will boil at 
.
Explanation:
According to clausius-clapeyron equation for liquid-vapor equilibrium:
![ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%3D%5Cfrac%7B-%5CDelta%20H_%7Bvap%7D%5E%7B0%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where, 
and 
are vapor pressures of liquid at 
(in kelvin) and 
(in kelvin) temperatures respectively.
Here, = 760.0 mm Hg, = 373 K, = 314.0 mm Hg
Plug-in all the given values in the above equation:
![ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B314.0%7D%7B760.0%7D%29%3D%5Cfrac%7B-40.7%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
or, 
So, 
Hence, at base camp, water will boil at
Answer:
P₂ = 299.11 KPa
Explanation:
Given data:
Initial volume = 600 mL
Initial pressure = 70.00 KPa
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 40°C (40+273 = 313 K)
Final volume = 150.0 mL
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL
P₂ = 13146000 KPa .mL. K /43950 K.mL
P₂ = 299.11 KPa
Answer:
Different Number of neutrons.
Answer:
17.6510 L
Explanation:
First we should get the number of moles of helium here by Boyle's law
PV=nRT
P=750/760= 0.9868 atm
T=25+273=298 kelvin
R= 0.08206
V= 20L
so
n=PV/RT
n=0.9868×20/0.08206×298
n=0.80707 mol
Then use the same law
V=0.80707×0.08206×263/0.9869=
17.6510L
SO THE VOLUME WILL BE 17.6510 L
