<u>Answer:</u> The volume when the pressure and temperature has changed is 1332.53 L
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
![P_1=0.950atm\\V_1=200L\\T_1=27^oC=[273+27]K=300K\\P_2=0.125atm\\V_2=?L\\T_2=-10^oC=[273-10]K=263K](https://tex.z-dn.net/?f=P_1%3D0.950atm%5C%5CV_1%3D200L%5C%5CT_1%3D27%5EoC%3D%5B273%2B27%5DK%3D300K%5C%5CP_2%3D0.125atm%5C%5CV_2%3D%3FL%5C%5CT_2%3D-10%5EoC%3D%5B273-10%5DK%3D263K)
Putting values in above equation, we get:
Hence, the volume when the pressure and temperature has changed is 1332.53 L
Answer:
790 g/s
Explanation:
The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.
The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;
Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].
Where g = wind direction, y and h are the distance in metres.
Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;
Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.
Emission rate of pollutant,E = 790 g/s.
Answer:
The new pressure at constant volume is 1066.56 kPa
Explanation:
Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.
Mathematically, P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
I believe your answer would be the first one
hope this helps
