IIT-JEE 2011 · Paper 1 · Physics Q25

PhysicsSingle correct+3 / -1Question 25 of 69

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be $T_1$ , the work done in the process is

\dfrac98RT_1

\dfrac32RT_1

\dfrac{15}8RT_1

\dfrac92RT_1

Answer and worked solution

Official answer: A

Moles. 5.6 L at STP: $n=\frac14$ mol of a monatomic gas ( $\gamma=\frac53$ ).

Final temperature. $T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=T_1\cdot8^{2/3}=4T_1$ .

Work on the gas. $|W|=nC_v\Delta T=\frac14\cdot\frac32R\cdot3T_1=\frac98RT_1$ .

(A)

Topics:Work done by a gas Thermodynamic processes

IIT-JEE 2011 · Paper 1, Physics question 25