One mole of an ideal diatomic gas expands fro | vedclass

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

Question

(B) For an isothermal process,the work done is given by $W_{	ext{isothermal}} = nRT \ln \left(\frac{V_2}{V_1}ight)$.Given $n = 1$,$T = 27^{\circ} C

Vedclass · Accepted Answer

$-56$

Vedclass · Answer

(B) For an isothermal process,the work done is given by $W_{	ext{isothermal}} = nRT \ln \left(\frac{V_2}{V_1}ight)$.Given $n = 1$,$T = 27^{\circ} C = 300 \ K$,and $V_2/V_1 = 2$,we have $W = 1 \cdot R \cdot 300 \cdot \ln(2) \approx 300R(0.693)$.For an adiabatic process,the work done is $W_{	ext{adiabatic}} = \frac{nR(T_1 - T_2)}{\gamma - 1}$.For a diatomic gas,$\gamma = 1.4$.Given $W_{	ext{isothermal}} = W_{	ext{adiabatic}}$,we equate the two expressions:$\frac{1 \cdot R(300 - T_{	ext{final}})}{1.4 - 1} = 300R(0.693)$.$\frac{300 - T_{	ext{final}}}{0.4} = 300(0.693)$.$300 - T_{	ext{final}} = 0.4 \cdot 300 \cdot 0.693 = 83.16$.$T_{	ext{final}} = 300 - 83.16 = 216.84 \ K$.Converting to Celsius: $T(^{\circ} C) = 216.84 - 273.15 \approx -56.31^{\circ} C$.Thus,the final temperature is close to $-56^{\circ} C$.

Similar Questions

In thermodynamics,heat and work are

$1$ mole of rigid diatomic gas performs a work of $Q/5$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{xR}{8}.$ The value of $x$ is $\ldots \ldots \ldots .$ $[R =$ universal gas constant $]$

$A$ gas expands such that its initial and final temperatures are equal. Also,the process followed by the gas traces a straight line on the $P-V$ diagram:

An ideal gas undergoes a thermodynamic cycle as shown in the figure. Which of the following graphs represents the same cycle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module