If a gas has a volume of 400.0 cm cubed at a pressure of 755 mmHg and a temperature of 25 degrees celsius, what would be its pressure (in atm) at 85 degrees celsius if the volume is reduced to 50.0 cm cubed?
(P1V1)/T1 = (P2V2)/T2
Is this correct?
(755 mmHg)(400.0cm^3)/298 K = (P2)(50.0cm^3)/(358 k)
looks good to me.
What volume of 1.60 M SrCl2is needed to prepare 525mL of 5.00 SrCl2
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Let's denote the initial conditions as follows:
P1 = 755 mmHg
V1 = 400.0 cm^3
T1 = 25 degrees Celsius
And the final conditions as:
P2 = ?
V2 = 50.0 cm^3
T2 = 85 degrees Celsius
Now let's substitute the known values into the equation and solve for P2:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(755 mmHg * 400.0 cm^3) / (25 + 273.15 K) = (P2 * 50.0 cm^3) / (85 + 273.15 K)
To convert temperatures from Celsius to Kelvin, we add 273.15 to the Celsius value.
Simplifying the equation:
(755 * 400.0) / (25 + 273.15) = (P2 * 50.0) / (85 + 273.15)
Calculating numerator:
302,000 / 298.15 = (P2 * 50.0) / (85 + 273.15)
Simplifying the expression further:
1013.25 mmHg = (P2 * 50.0) / 358.15
Cross multiply:
(1013.25 mmHg) * (358.15) = P2 * 50.0
363338.21875 mmHg cm^3 = P2 * 50.0 cm^3
Divide both sides by 50.0 cm^3:
P2 = 363338.21875 mmHg / 50.0 cm^3
Dividing both sides by the conversion factor, 760 mmHg = 1 atm:
P2 = (363338.21875 mmHg / 50.0 cm^3) * (1 atm / 760 mmHg)
Simplifying:
P2 = 4.78 atm
Therefore, the pressure of the gas at 85 degrees Celsius and a volume of 50.0 cm^3 would be approximately 4.78 atm.