One mole of an ideal monoatomic gas is subjec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The process consists of three steps:
$1 \rightarrow 2$: Isobaric expansion at $P = 1.0 \ bar$ from $V = 20 \ L$ to $V = 40 \ L$.
$W_{1 \rightarr

Vedclass · Accepted Answer

$620$

Vedclass · Answer

(A) The process consists of three steps:
$1 ightarrow 2$: Isobaric expansion at $P = 1.0 \ bar$ from $V = 20 \ L$ to $V = 40 \ L$.
$W_{1 ightarrow 2} = -P \Delta V = -1.0 \ bar 	imes (40 - 20) \ L = -20 \ bar \ L$.
$2 ightarrow 3$: Isochoric cooling at $V = 40 \ L$ from $P = 1.0 \ bar$ to $P = 0.5 \ bar$.
$W_{2 ightarrow 3} = 0 \ J$ (since $\Delta V = 0$).
$3 ightarrow 1$: Isothermal compression at $T$ (constant) from $V = 40 \ L$ to $V = 20 \ L$.
$W_{3 ightarrow 1} = -nRT \ln(\frac{V_1}{V_3}) = -P_3 V_3 \ln(\frac{V_1}{V_3})$.
Given $P_3 = 0.5 \ bar$ and $V_3 = 40 \ L$,$P_3 V_3 = 0.5 	imes 40 = 20 \ bar \ L$.
$W_{3 ightarrow 1} = -20 \ln(\frac{20}{40}) = -20 \ln(0.5) = 20 \ln 2$.
Using $\ln 2 = \log_{10} 2 	imes \ln 10 = 0.3 	imes 2.3 = 0.69$.
$W_{3 ightarrow 1} = 20 	imes 0.69 = 13.8 \ bar \ L$.
$	ext{Total work } W = W_{1 	o 2} + W_{2 	o 3} + W_{3 	o 1} = -20 + 0 + 13.8 = -6.2 	ext{ bar L}$Magnitude $|W| = 6.2 \ bar \ L$.
Since $1 \ bar \ L = 100 \ J$,$|W| = 6.2 	imes 100 = 620 \ J$.

One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is $...........J$ (nearest integer). Given: $\log 2=0.3, \ln 10=2.3$

Similar Questions

$A$ gas at $10 \, atm$ pressure and $300 \, K$ temperature undergoes adiabatic reversible expansion to $5 \, atm$ pressure and $290 \, K$ temperature. The molar heat capacity $C_v$ of the gas in $cal \, mol^{-1} \, K^{-1}$ is:

The following reaction occurs in an automobile: $2C_8H_{18(g)} + 25O_{2(g)} \to 16CO_{2(g)} + 18H_2O_{(g)}$. The signs of $\Delta H$,$\Delta S$,and $\Delta G$ would be:

For irreversible expansion of an ideal gas under isothermal condition,the correct option is :

An isolated box,equally partitioned,contains two ideal gases $A$ and $B$ as shown in the figure. When the partition is removed,the gases mix. The changes in enthalpy $(\Delta H)$ and entropy $(\Delta S)$ in the process,respectively,are

The enthalpy of formation of $H_2O$ is $-68 \ k.cal/mol$. Calculate the enthalpy of formation of $OH^-$. Given that the enthalpy of neutralization of $H^+$ and $OH^-$ is $-13.7 \ k.cal/mol$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module