Question

A compression, at a constant pressure of 140 kPa, is performed on 4.0 moles of an ideal monatomic gas (Cv = 3/2 R). The compression reduces the volume of the gas from 0.26 m^3 to 0.12 m^3. The change in the internal energy of the gas, in kJ is _____? ("^3" means to the power of 3)

Answer 1

Answer: The change in internal energy of the gas is 29.414 kJ.

Explanation:

To calculate the temperature of the gas at different volumes, we use ideal gas equation:

$PV=nRT$

• When volume = $0.26m^3$

We are given:

Conversion used:  $1m^3=1000L$

$P=140kPa\\V=0.23m^3=260L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}$

Putting values in above equation:

$140kPa\times 260L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_i\\\\T_i=1095.06K$

• When volume = $0.12m^3$

We are given:

$P=140kPa\\V=0.12m^3=120L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}$

Putting values in above equation:

$140kPa\times 120L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_f\\\\T_f=505.41K$

• To calculate the change in internal energy, we use the equation:

$\Delta U=nC_v\Delta T=nC_v(T_f-T_i)$

where,

$\Delta U$ = change in internal energy = ?

n = number of moles = 4.0 mol

$C_v$ = heat capacity at constant volume = $\frac{3}{2}R$

$T_f$ = final temperature = 1095.06 K

$T_i$ = initial temperature = 505.41 K

Putting values in above equation, we get:

$\Delta U=4\times \frac{3}{2}\times 8.314J/K.mol\times (505.41-1095.06)\\\\\Delta U=29414.1J$

Converting this value in kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, $29414.1J=\frac{1kJ}{1000J}\times 29414.1J=29.414kJ$

Hence, the change in internal energy of the gas is 29.414 kJ.