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[edit] 21.

The pressure exerted on a 240. mL sample of hydrogen gas at a constant temperature is increased from 0.428 atm to 0.724 atm. What will the final volume of the sample be?

(240.mLH2)(0.428atm) = (0.724atm)(x)
102.72atmmLH2 = 0.724xatm
141.878mLH2 = x
Vf = 142mLH2

[edit] 26.

A sample of air has a volume of 140.0mL at 67°C. At what temperature will its volume be 50.0mL at constant pressure?

67°C + 273.15 = 340.15 K → 340. K
\frac{140.0mL}{340.K}=\frac{50.0mL}{x}
140.0xmL = 17000mLK
x = 121.429K
121.429 K - 273.15 = -151.721°C
Tf=-151°C

[edit] 27.

At standard temperature, a gas has a volume of 275mL. The temperature is then increased to 130.°C. and the pressure is held constant. What is the new volume?

0°C + 273.15 = 273.15 K → 273 K
130.°C + 273.15 = 400.15 K → 400 K
\frac{275mL}{273K}=\frac{x}{400K}
110000mLK = 273xK
402.930mL = x
Vf = 403mL

[edit] 28.

A sample of hydrogen at 47°C exerts a pressure of 0.329 atm. The gas is heated to 77°C at constant volume. What will its new pressure be?

47°C + 273.15 = 320.15 K → 320. K
77°C + 273.15 K = 350.15 K → 350. K
\frac{0.329atm}{320.K}=\frac{x}{350.K}
115.15atmK = 320xK
0.360atm = x
Pf = 0.360atm

[edit] 31.

A sample of gas at 47°C and 1.03 atm occupies a volume of 2.20 mL. What volume would this gas occupy at 107°C and 0.789 atm?

47°C + 273.15 = 320.15 K → 320. K
107°C + 273.15 = 380.15 K → 380. K
\frac{(1.03atm)(2.20mL)}{320.K}=\frac{(0.789atm)(x)}{380.K}
\frac{2.266atm mL}{320K}=\frac{0.789x atm}{380K}
861.08atmmLK = 252.48xatmK
3.410mL = x
Vf = 3.41mL

[edit] 34.

A sample of oxygen at 40.°C occupies 820. mL. If this sample occupies 1250 mL at 60.°C and 1.40 atm, what was its original pressure?

40.°C + 273.15 = 313.15 K → 313. K
60°C + 273.15 = 333.15 K → 333. K
\frac{(x)(820.mL)}{313.K}=\frac{(1.40atm)(1250mL)}{333.K}
\frac{820x mL}{313K}=\frac{1750atm mL}{333K}
273060xmLK = 547750atmmLK
x = 2.006atm
Pi = 2.01atm