Question

Calculate the work done in compressing adiabatically 3kg of helium (He) to one fifth of its original volume if it is initially at 13°C. Find the change in internal energy of the gas resulting from the compression. (cp/cv for monatomic gases is 1.667; gas constant for helium is 2079 K^-1 kg^-1)

Answer 1

Answer:

Work done,$w=5.12\times10^{6}\ \rm J$

change in internal Energy ,$\Delta U=5.12\times10^6\ \rm J$

Explanation:

Given:

• Mass of helium gas $m=3\ \rm kg$
• initial temperature $T_i=286\ \rm K$

Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings

$T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K$

Let n be the number of moles of Helium given by

$n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3$

Work done in Adiabatic process

Let W be the work done

$W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J$

The Internal Energy change in any Process is given by

Let $\Delta U$ be the change in internal Energy

$\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J$