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2 years ago

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What is the effect of the following on the volume of 1 mol of an ideal gas? The temperature changes from 305 K to 32°C and the p

ressure changes from 2 atm to 101 kPa.
A. Volume increases by 200%
B. Volume decreases to 50% of its original value
C. Volume decreases by 100%
D. Volume increases by 100%
E. Volume does not change

1 answer:

ryzh [129]2 years ago

7 0

Answer:

The volume increases by 100%.

Explanation:

<u>Step 1:</u> Data given

Number of moles ideal gas = 1 mol

Initial temperature = 305 K

Final temperature = 32°C + 273.15 = 305.15 K

Initial pressure = 2 atm

final pressure = 101 kPa = 0.996792 atm

R = gasconstant = doesn't change

V1 = initial volume

V2= the final volume

<u>Step 2: </u>Calculate volume of original gas

P*V = n*R*T

(P*V)/ T = constante

(P1 * V1) / T1 = (P2 * V2)/ T2

In this situation we have:

(2atm * V1)/ 305 = (0.996792 *V2) / 305.15

0.006557*V1 = 0.003266*V2

V2 = 2*V1

We see that the final volume is twice the initial volume. So the volume gets doubled. The volume increases by 100%.

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Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

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2 years ago