120 degres celcious
Raising the temperature of a gas will only increase the pressure if the volume is conserved, decreased, or (only sometimes) increased. The equation relating all three of these variables is:PV = nRT, orPressure(in pascales) x Volume(in m3) = (number of moles of gas) x 8.3145(J/mol*K) x Temperature(in degrees Kelvin)We can figure out the answer to your question by using some arbitrary numbers in this equation: we will say that the original Pressure is 101325 pascales (1atm), Volume is 2m3, T is 300oK, and the number of moles of gas is 81.24 moles.This gives us the equation for the original system:(101325 pascales)(2m3)=(8.3145J/mol*K)(81.24mol)(300oK)Let's now increase the temperature by 50oK, to 350oK. We will also conserve the original volume, and see if the pressure increases. This gives us the new equation:(X pascales)(2m3)=(8.3145J/mol*K)(81.24mol)(350oK).Since both n and R are constants when considering the same system, we do not need to investigate what will happen when these two are changed (since a change in these two is practically infeasable).If we solve for Pressure algebraically, we find that:X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(2m3) = 118207 pascales.Since our new pressure, 118207 pascales, is larger than the orgininal pressure, 101325 pascales, we can see that an increase in temperature results in an increase in pressure if volume is conserved.Additionally, we can investigate what happens when the volume is decreased by 1m3 (from 2m3 to 1m3) along with an increase in temperature by 50oK, to 350oK. We will be comparing this new system to the old system already established in the previous part of this problem. This gives us an equation for the new system:(X pascales)(1m3)=(8.3145J/mol*K)(81.24mol)(350oK)If we solve for Pressure algebraically, we find that:X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(1m3) = 236414 pascales.Since our new pressure, 236414 pascales, is larger than the orgininal pressure, 101325 pascales, we can see that an increase in temperature results in an increase in pressure if volume is decreased.But: what happens if the Volume is increased? Let's consider a situation where an increase in temperature by 50oK, to 350oK is accompanied by a decrease in volume by 1m3, from 2m3 to 3m3. This gives us the equation:(X pascales)(3m3)=(8.3145J/mol*K)(81.24mol)(350oK)If we solve for Pressure algebraically, we find that:X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(3m3) = 78804 pascales.Since our new pressure, 78804 pascales, is smaller than the orgininal pressure, 101325 pascales, we can see that an increase in temperature DOES NOT always result in an increase in pressure if volume is increased.As a general rule, we can therefore conclude that an increase in volume does not always result in an increase in pressure if the temperature is increased. However, there are marginal situations in which this CAN happen: if, for example, the volume is increased by 0.0000001m3, along with an increase in temperature to 350oK, we find that solving for the Pressure gives us a value of 118207 pascales, which is larger than our original pressure, 101325 pascales. So, we can see that an increase in temperature CAN SOMETIMES result in an increase in pressure if volume is increased.An increase in volume only increases pressure in a system if the increase in volume is significantly small when compared to the increase in temperature.
The volume of a cube equals its side length cubed. In this case it is 3m3 which gives a volume of 27m3
The volume of a cube equals its side length cubed. In this case it is 3m3 which gives a volume of 27m3
The formula for density is mass/volume.mass = 10 gramsvolume = 3m3density = 10 g /3m3 = 3.333 g/m3
You can write 3m3
608.05 sq. m
Density of ice is 917 kg/m3 Volume is mass/density is (1kg)/(917 kg/m3)=1.09x10-3m3 and 1 litre=1m3 then 1 kilogram of ice is 1.09x10-3 litres.
1mm of rain is actually water volume of 1lt per 1m2 of surface, 1mm=1lt/1m2 explanation: Simply convert 1lt to 1000 ml. Since 1ml=1cm3 and 1cm=0.01m=10-2m then 1cm3 = 10-6m3 this means that 1lt=1000ml=1000cm3 = 1000 x 10-6m3= 10-3m3 So 1lt/1m2 from the above is 10-3m3/1m2= 10-3m = 1mm. I hope that it is not complicated.
The density is 20 kg/m3
You cannot answer this question mathematically as you do not know the length of the hole, if it is a square hole (i.e. 1m long) then there was 3m3 of dirt.However if it is a lateral thinking question then the answer is none. It is a hole, therefore it is empty
this is going to be very small. 1 nanometre is 10-9 metres. 1 nanometre cubed is 10-9(3) which is 10-27 metres cubed. A litre is 1000 cm3. there are 100cm in a metre, so in a centimetre its 10-2 metres, to the power of 3, there are 10-6m3 in a ml. in 1 litre there are 10-3m3. therefore, there are 1024nm3 in a litre (by my calculation)
There is no conversion. Cubic feet is always cubic feet. Standard = normal. This is not true, standard cubic feet is a cubic feet of a substance (typically a gas) at 60 °F and 1 ATM pressure where as normal cubic feet would be the volume of a substance (again typically a gas) at 0 °C and 1 ATM pressure. It follows that to convert from one to another, an estimate can be made using the ideal gas equation. If you need any further instruction let me know. It's a secret, so write to him. It's not a secret Jens I just didn't want to extensively talk about the ideal gas law since lots of people already know about it.