Question

Two moles of an ideal gas $\left(C_v=\frac{5}{2}R\right)$ was compressed adiabatically against constant pressure of $2atm$, which was initially at $350K$ and $1atm$ pressure. The work involve in the process is equal to?

• A. $250R$
• B. $300R$
• C. $400R$
• D. $500R$

Answer 1

The correct option is D $500R$

For reversible adiabatic process,

$W=-P_{ext}[\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1}]$

$P_2=P_{ext}=2atm$

$P_1=1atm$ $T_1=300K$

$W=-\left(2atm\right)[\frac{2\left(R\right).T_2}{2atm}-\frac{2R\left(350\right)}{1atm}]$

and $W=2C_V(T_2-350)=2\times \frac{5}{2}R(T_2-350)$

$5R(T_2-350)=(750R-2R{T}_2)$

$5T_2-1750=1400-2T_2$

$7T_2=3150$ $T_2=450K$

$W=2\times C_V(450-350)$

$=2\times \frac{5}{2}R\times \left(100\right)=500R$

$W=W_{AB}+W_{BC}+W_{CD}$
$=-P_0(V_B-V_A)=-nR{T}_{B}ln\left[\frac{V_C}{V_B}\right]-\frac{P_0}{2}(V_D-V_C)$

$W=-P_0(2V_0-V_0)-2P_0{V}_0$ ln

$\left[\frac{4V_0}{2V_0}\right]-\frac{P_0}{2}(2V_0-4V_0)$

$W=-2P_0{V}_{0}ln2$ and $q=-W(\because \Delta U=0)$

$q=2P_0{V}_{0}ln2$