Answer:
1) 1.52 atm.
2) 647.85 K.
3) 20.56 L.
4) 1.513 mole.
5) 254.22 K = -18.77 °C.
Explanation:
- In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
- Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).
1) In this point; n, R, and T are constants and the variables are P and V.
P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.
<u><em>In our problem:</em></u>
P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.
P2 = 5.7 atm and V2 = 12.0 L.
Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.
2) In this case, n and R are the constants and the variables are P, V, and T.
P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.
<u><em>In our problem:</em></u>
P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.
P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>
Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.
3) In this case, P, n and R are the constants and the variables are V, and T.
V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.
<u><em>In our problem:</em></u>
V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.
V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.
Herein, there is no necessary to convert T into K.
Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.
4) We can get the number of moles that will fill the container from: n = PV/RT.
P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.
V = 16.0 L.
R = 0.082 L.atm/mol.K.
T = 45 °C + 273 = 318 K.
Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.
5) In this case, V, n and R are the constants and the variables are P, and T.
P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.
<u><em>In our problem:</em></u>
P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.
P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.
Herein, there is no necessary to convert P into atm.
Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.
