Potential energy

When the burner in a hot air balloon is turned on, the temperature of the air in the balloon ________ causing its volume to ____

raketka [301]

B) increases, expand, rises

4 0

4 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

A blank is a coil of wire that produces a magnetic field when it carries an electric current

Lostsunrise [7]

A spring - also an electromagnet. Also know as a slinky!

7 0

4 years ago

Read 2 more answers

How many moles of KCL is in a 3M solution of KCL

netineya [11]

Answer:

223.6539

Explanation:

1 mole is equal to 1 moles KCl, or 74.5513 grams

4 0

3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago