The initial pressure is halved. Use Boyle's law that relates pressure & volume at a constant temperature.
P1V1 = P2V2
In this case the V1(initial volume) is doubled so V2 = 2V1
P2 = P1V1/V2 = P1V1/2V1
P2 = (1/2)*P1
For an ideal gas, per the Ideal Gas Law, pV = nRT, if temperature doubles, then pressure doubles.
pv=nrt
if temp and pressure double, the equation is sound, and volume remains constant.
PV=nRT, so P=nRT/V for an ideal gas, so if the volume is doubled and the temperature is halved, the pressure remains constant.
Pressure doubles too. Temperature and pressure share a direct relationship (this is called Gay-Lussac's Law.)
Assuming the volume remains constant, doubling the temperature will increase the pressure by 2 times, i.e. it will also double.
Assuming the volume of the container and the amount of product in it remain constant, if temperature doubles pressure will also double.
The volume is decreased to 1/2.
The pressure is also doubled.
EXPLOSSION
.. thenEITHER the pressure is halved for the same amount (moles) of gas,ORthe amount (moles) of gas is doubled at the same pressure,ORany valid combination of these possibillities.
Increasing the the pressure the volume decrease.The law of Boyle and Mariotte: P.V= k
At normal pressure, between 161.4 K and 165.1 K.
200
When the pressure of a gas is increased, the volume of the gas is decreased. When the pressure of the gas is decreased, the volume increases
The pressure is doubled (up to SP=760 Torr) and the temperature is halved (down to ST=273 K) as long as the volume (and amount of matter) is UNchanged!
Increasing the the pressure the volume decrease.The law of Boyle and Mariotte: P.V= k
Increasing the the pressure the volume decrease.The law of Boyle and Mariotte: P.V= k
Increasing the the pressure the volume decrease.The law of Boyle and Mariotte: P.V= k
.. thenEITHER the pressure is halved for the same amount (moles) of gas,ORthe amount (moles) of gas is doubled at the same pressure,ORany valid combination of these possibillities.
Increasing the the pressure the volume decrease.The law of Boyle and Mariotte: P.V= k
It becomes double as K=Q/t×L/A(T2-T1) so if the thickness (L) of an object is doubled the thermal conductivity will be doubled as thermal conductivity is directly proportional to the thickness/L of the object K=L K=2L,K=2 two times
At stp or standard temperature and pressure, we have pressure = 100kPa and temperature=273.15 K
60kpa
At normal pressure, between 161.4 K and 165.1 K.
273 K (0° Celsius) and 1 ATM pressure
At normal pressure, it is 161.4 K.