Physics, 13.02.2020 23:54 TheOverlordOfWhales

A) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3 closed container. If the gas goes through an isochoric process to twice the initial temperature, what is the new pressure of the gas in Pa?

b) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3closed container. If the gas goes through an isothermal process to 3.6 m3, what is the new pressure of the gas in Pa?

c) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3 closed container. If the gas goes through an isobaric process to 3.6 m3, what is the new temperature of the gas in Kelvin?

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Answer from: angelinad02

a) $P_f = \frac{nRT_f}{V_f}=\frac{1.3 mol * 0.082 \frac{atm L}{mol K} 800 K}{1200 L}= 0.071 atm = 7200. 83 Pa$

b) $P_f = \frac{nRT_f}{V_f}= \frac{1.3 mol * 0.082 \frac{atm L}{mol K} 400 K}{3600L}= 0.0118 atm = 1200.138 Pa$

c) $T_f = \frac{P_f V_f}{nR}= \frac{0.0355 atm *3600 L}{1.3 mol 0.082 \frac{atm L}{mol K}}= 1200 K$

Explanation:

a) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3 closed container. If the gas goes through an isochoric process to twice the initial temperature, what is the new pressure of the gas in Pa?

For this case we have the initial conditions given:

n = 1.3 mol, T_i = 400 K, V_i = 1.2 m^3 = 1200 L = V_f

The volume not changes since we have an isochoric process.

For this case we know that the final temperature would be T_f = 2T_i = 800 K , V_f = 1200 L n = 1.3 mol and we can find the final pressure like this:

$P_f = \frac{nRT_f}{V_f}=\frac{1.3 mol * 0.082 \frac{atm L}{mol K} 800 K}{1200 L}= 0.071 atm = 7200. 83 Pa$

b) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3closed container. If the gas goes through an isothermal process to 3.6 m3, what is the new pressure of the gas in Pa?

For this case we have the following info:

T_i =400 K = T_f since the process is isothermal

n = 1.3 mol V_i = 1.2 m^3 = 1200 L , V_f = 3.6 m^3 = 3600 L

From the ideal gas law we have:

PV = nRT

And if solve for the final pressure we got:

$P_f = \frac{nRT_f}{V_f}= \frac{1.3 mol * 0.082 \frac{atm L}{mol K} 400 K}{3600L}= 0.0118 atm = 1200.138 Pa$

c) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m3 closed container. If the gas goes through an isobaric process to 3.6 m3, what is the new temperature of the gas in Kelvin?

For this case we have the following info:

n = 1.3 mol, T_i = 400 K, V_i = 1.2 m^3 = 1200 L, V_f = 3.6 m^2 = 3600 L

From the initial condition we can find the initial pressure using the ideal gas law:

PV = nRT

$P_i = \frac{nRT_i}{V_i}=\frac{1.3 mol * 0.082 \frac{atm L}{mol K} 400 K}{1200 L}= 0.0355 atm= 3600.415 Pa$

And we know that the pressure not change since the process is isobaric so then P_i = P_f = 0.0355 atm and we can find the final temperature like this:

$T_f = \frac{P_f V_f}{nR}= \frac{0.0355 atm *3600 L}{1.3 mol 0.082 \frac{atm L}{mol K}}= 1200 K$

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