Question

A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J

Answer 1

Answer:

a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy is -18279.78 J

c) heat lost by the gas is zero or 0

d) the work done on the gas is -18279.78 J

Explanation:

Given the data in the question;

P$_i$ = 1 atm = 101325 pascal

P$_f$ = ?

V$_i$ = 0.1550 m³

V$_f$ = 0.742 m³

we know that for an adiabatic process  γ = 1.4

P$_i$V$_i^Y$ = P$_f$V$_f^Y$

P$_f$ = P$_i$$($ V$_i$ / V$_f$ $)^Y$

we substitute

P$_f$ = 1 × $($ 0.1550  / 0.742  $)^{1.4$

= $($ 0.2088948787 $)^{1.4$

= 0.11166 atm

a) the initial and final temperatures

Initial temperature

T$_i$ = P$_i$V$_i$ / nR

given that n = 3.10 mol

= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )

= 15705.375 / 25.7734

T$_i$  = 609.4 K

Final temperature

T$_f$ = P$_f$V$_f$ / nR

= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )

= 8394.95 / 25.7734

= 325.7 K

Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy

ΔE$_{int$ = nC$_v$ΔT

here, C$_v$ = ( 5/2 )R

ΔE$_{int$ = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

= -18279.78 J

Therefore, the change in internal energy is -18279.78 J

c) the heat lost by the gas

Since its an adiabatic process,

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  the work done on the gas

W = ΔE$_{int$ - Q

= -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J

Answer 2

a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.

b) The change in internal energy is -18279.78 J.

c) The heat lost by the gas is zero.

d) The work done on the gas is -18279.78 J.

Given data:

The moles of sample is, n = 3.10 mol.

The initial volume of sample is, $V_{1}=0.1550 \;\rm m^{3}$.

The final volume of sample is, $V_{2}=0.742 \;\rm m^{3}$.

The initial pressure of the sample is, $P_{1}=1.00 \;\rm atm$.

(a)

We know that the relation between the pressure and volume for an adiabatic process is as follows,

$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$

Here, $\gamma$  is a adiabatic index. And for air, its value is 1.41.

Solving as,

$P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm$

Now, calculate the final temperature using the ideal gas equation as,

$P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K$

Similarly, calculate the initial temperature as,

$P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K$

Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.

(b)

The change in internal energy is given as,

ΔE = nCΔT

here, C = ( 5/2 )R

ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

      = -18279.78 J

Therefore, the change in internal energy is -18279.78 J.

c)

The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.  

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  

The work done on the gas

W = ΔE - Q

W = -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J.

Learn more about the adiabatic process here:

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