Gas is confined in a tank at a pressure of 1.14×108Pa and a temperature of 13.4◦C.?
2018-06-28 12:10 am
Gas is confined in a tank at a pressure of 1.14×10^8 Pa and a temperature of 13.4◦C. Half of the gas is withdrawn and the temperature is raised to 84.1◦C.Find the new pressure in the tank. Answer in units of Pa.
回答 (3)
2018-06-28 1:31 am
Initial: P₁ = 1.14 × 10⁸ Pa, T₁ = (273.2 + 13.4) K = 286.6 K, n₁ = n
New: P₂ = ? Pa, T₂ = (273.2 + 84.1) K = 357.3 K, n₂ = 0.5n
Gas law: PV = nRT
At constant V: nT/P = V/R = constant
Hence, n₁T₁/P₁ = n₂T₂/P₂
New pressure, P₂
= P₁ × (n₂/n₁) × (T₂/T₁)
= (1.14 × 10⁸ Pa) × (0.5n/n) × (357.3/286.6)
= 7.11 × 10⁷ Pa
New: P₂ = ? Pa, T₂ = (273.2 + 84.1) K = 357.3 K, n₂ = 0.5n
Gas law: PV = nRT
At constant V: nT/P = V/R = constant
Hence, n₁T₁/P₁ = n₂T₂/P₂
New pressure, P₂
= P₁ × (n₂/n₁) × (T₂/T₁)
= (1.14 × 10⁸ Pa) × (0.5n/n) × (357.3/286.6)
= 7.11 × 10⁷ Pa
2018-06-28 12:21 am
initial:
n = PV/RT = (1.14e8)(V) / (R)(273.16+13.4)
final
n/2 = PV/RT = PV / (R)(273.16+84.1)
n = 2PV / (R)(273.16+84.1)
set the n's equal
(1.14e8)(V) / (R)(273.16+13.4) = 2PV / (R)(273.16+84.1)
cancel P and R
(1.14e8) / (273.16+13.4) = 2P / (273.16+84.1)
2P = (1.14e8)(273.16+84.1) / (273.16+13.4)
2P = (1.14e8)(357.26) / 286.56
P = 71063000 Pa
Ideal gas law
PV = nRT
n = number of moles
R = gas constant = 0.08206 (atm∙L)/(mol∙K)
T = temperature in kelvins
P = absolute pressure in atm
V = volume in liters
n = PV/RT = (1.14e8)(V) / (R)(273.16+13.4)
final
n/2 = PV/RT = PV / (R)(273.16+84.1)
n = 2PV / (R)(273.16+84.1)
set the n's equal
(1.14e8)(V) / (R)(273.16+13.4) = 2PV / (R)(273.16+84.1)
cancel P and R
(1.14e8) / (273.16+13.4) = 2P / (273.16+84.1)
2P = (1.14e8)(273.16+84.1) / (273.16+13.4)
2P = (1.14e8)(357.26) / 286.56
P = 71063000 Pa
Ideal gas law
PV = nRT
n = number of moles
R = gas constant = 0.08206 (atm∙L)/(mol∙K)
T = temperature in kelvins
P = absolute pressure in atm
V = volume in liters
2018-06-28 5:12 am
We will use the following equation.
P * V = n * R * T
In this problem, the volume of the gas is constant. R is also constant. To solve this problem, the temperatures must be converted from ˚C to ˚K by adding 273˚.
T1 = 13.4 + 273 = 286.4
T2 = 84.1 + 273 = 357.1
The pressure must be in atmospheres. One atmosphere is 1.013 * 10^5 Pa. To convert the pressure in Pascals to atmospheres, divide by this number.
P1 = 1.14 * 10^8 ÷ 1.013 * 10^5
The pressure is approximately 1,422 atmospheres.
(1.14 * 10^8 ÷ 1.013 * 10^5) * V = n * R * 286.4
Let’s solve this equation for n.
n = (1.14 * 10^8 ÷ 1.013 * 10^5) * V ÷ (R * 286.4)
(P2 ÷ 1.013 * 10^5) * V = n/2 * R* 357.1
Let’s solve this equation for n.
n = 2 * (P2 ÷ 1.013 * 10^5) * V ÷ (R * 357.1)
Set these two equations equal to each other and solve for P2.
2 * (P2 ÷ 1.013 * 10^5) * V ÷ (R * 357.1) = (1.14 * 10^8 ÷ 1.013 * 10^5) * V ÷ (R * 286.4)
Multiply both sides of this equation by 1.013 * 10^5.
2 * P2 * V ÷ (R * 357.1) = 1.14 * 10^8 * V ÷ (R * 286.4)
Divide both sides of equation by V ÷ R.
2 * P2 ÷ 357.1 = 1.14 * 10^8 ÷ 286.4)
Multiply both sides of this equation by 357.1 * 286.4.
572.8 * P2 = 4.07094 * 10^10
P2 = 4.07094 * 10^10 ÷ 572.8
The new pressure is approximately 7.11 * 10^7 atmospheres. To convert to Pascals, multiply the pressure in atmospheres by 1.013 * 10^5.
P2 = 1.013 * 10^5 * 4.07094 * 10^10 ÷ 572.8
The new pressure is approximately 7.2 * 10^12 atm.
P * V = n * R * T
In this problem, the volume of the gas is constant. R is also constant. To solve this problem, the temperatures must be converted from ˚C to ˚K by adding 273˚.
T1 = 13.4 + 273 = 286.4
T2 = 84.1 + 273 = 357.1
The pressure must be in atmospheres. One atmosphere is 1.013 * 10^5 Pa. To convert the pressure in Pascals to atmospheres, divide by this number.
P1 = 1.14 * 10^8 ÷ 1.013 * 10^5
The pressure is approximately 1,422 atmospheres.
(1.14 * 10^8 ÷ 1.013 * 10^5) * V = n * R * 286.4
Let’s solve this equation for n.
n = (1.14 * 10^8 ÷ 1.013 * 10^5) * V ÷ (R * 286.4)
(P2 ÷ 1.013 * 10^5) * V = n/2 * R* 357.1
Let’s solve this equation for n.
n = 2 * (P2 ÷ 1.013 * 10^5) * V ÷ (R * 357.1)
Set these two equations equal to each other and solve for P2.
2 * (P2 ÷ 1.013 * 10^5) * V ÷ (R * 357.1) = (1.14 * 10^8 ÷ 1.013 * 10^5) * V ÷ (R * 286.4)
Multiply both sides of this equation by 1.013 * 10^5.
2 * P2 * V ÷ (R * 357.1) = 1.14 * 10^8 * V ÷ (R * 286.4)
Divide both sides of equation by V ÷ R.
2 * P2 ÷ 357.1 = 1.14 * 10^8 ÷ 286.4)
Multiply both sides of this equation by 357.1 * 286.4.
572.8 * P2 = 4.07094 * 10^10
P2 = 4.07094 * 10^10 ÷ 572.8
The new pressure is approximately 7.11 * 10^7 atmospheres. To convert to Pascals, multiply the pressure in atmospheres by 1.013 * 10^5.
P2 = 1.013 * 10^5 * 4.07094 * 10^10 ÷ 572.8
The new pressure is approximately 7.2 * 10^12 atm.
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