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2 years ago

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If a certain gas occupies a volume of 13 L when the applied pressure is 6.5 atm , find the pressure when the gas occupies a volu

me of 3.3 L

2 answers:

Arte-miy333 [17]2 years ago

7 0

Answer : The pressure when the gas occupies a volume of 3.3 L is, 25.6 atm.

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

$P\propto \frac{1}{V}$

or,

$PV=k$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of the gas = 6.5 atm

$P_2$ = final pressure of the gas = ?

$V_1$ = initial volume of the gas = 13 L

$V_2$ = final volume of the gas = 3.3 L

Now put all the given values in this formula, we get the final pressure of the gas.

$6.5atm\times 13L=P_2\times 3.3L$

$P_2=25.6atm$

Therefore, the pressure when the gas occupies a volume of 3.3 L is, 25.6 atm.

Sav [38]2 years ago

5 0

We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)

Solve for P2 on your calculator and that should get you to the answer.

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What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?

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2 years ago

Read 2 more answers

Ka, the acid dissociation constant, for an acid is 9 × 10^{−4} at room temperature. At this temperature, what is the approximate

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2 years ago

One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori

Igoryamba

Answer:

$P_2=0.398bar=39800Pa$

$T_2=118.7K\\$

$Q=-3729.9J$

$W=-61753.24J$

Δ $U_T=0J$

Δ $H_T=0J$

Explanation:

Hello,

At the first state, the molar volume is:

$v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3$

The volume in both the second and third state:

$v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3$

Now, as it is about an adiabatic process, one remembers the following relationships:

$PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4$

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

$P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa$

- And the temperature:

$T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\$

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

$Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J$

Then the total heat:

$Q=Q_1+Q_2=0-3729.9J=-3729.9J$

- The work for the first process is:

$W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J$

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

$W=W_1+W_2=-61753.24J+0J=-61753.24J$

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

$U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\$

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

$H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\$

Best regards.

8 0

2 years ago