Question

Two moles of an ideal gas $(C_v=\frac{5}{2}R)$ was compressed adiabatically against constant pressure of 2 atm. Which was initially at 350 K and 1 atm pressure. The work involve in the process is equal to:
Given: ln 350 = 5.857 and $e^{6.05}=426.65$

• A. 250 R
• B. 300 R
• C. 380 R
• D. 500 R

Answer 1

The correct option is C 380 R

For an adiabatic process,
$T^{\gamma }{P}^{1-\gamma }=$ constant
Since, $C_v=\frac{5R}{2}$. So, $\gamma$ = 1.4
$T_1^{\gamma }{P}_1^{1-\gamma }=$$T_2^{\gamma }{P}_2^{1-\gamma }$
${350}^{1.4}{1}^{1-1.4}=T_2^{1.4}{2}^{1-1.4}$
On taking log both sides, we get
$1.4ln350=1.4ln{T}_2-0.4ln2$
$T_2$ = 426.65 K
$W=n{C}_v(T_2-T_1)=2\times \frac{5R}{2}(426.65-350$)
W = 380R