An ideal gas is subjected to an isothermal ex | vedclass

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

Question

(D) For an isothermal process,the work done $BY$ the gas is given by $W_{by} = \int_{V_i}^{V_f} P \, dV$. Since $PV = nRT$,we have $P = nRT/V$. Integr

Vedclass · Accepted Answer

$W = -nRT \ln(P_f/P_i)$

Vedclass · Answer

(D) For an isothermal process,the work done $BY$ the gas is given by $W_{by} = \int_{V_i}^{V_f} P \, dV$. Since $PV = nRT$,we have $P = nRT/V$. Integrating this,$W_{by} = nRT \int_{V_i}^{V_f} \frac{1}{V} dV = nRT \ln(V_f/V_i)$. By Boyle's Law,$P_i V_i = P_f V_f$,so $V_f/V_i = P_i/P_f$. Thus,$W_{by} = nRT \ln(P_i/P_f) = -nRT \ln(P_f/P_i)$. The work done $ON$ the gas is the negative of the work done by the gas: $W_{on} = -W_{by} = -nRT \ln(V_f/V_i) = nRT \ln(P_f/P_i)$.

An ideal gas is subjected to an isothermal expansion such that its volume changes from $V_i$ to $V_f$ and pressure from $P_i$ to $P_f$. The work done on the gas is:

Similar Questions

One mole of an ideal gas expands from an initial volume of $10 \; L$ to a final volume of $20 \; L$ at a constant temperature of $300 \; K$. Calculate the work done by the gas in $J$. $(R = 8.31 \; J/mol \cdot K)$ (in $; J$)

Can two isothermal curves cut each other?

Can $PV = RT$ be described with an isothermal or an adiabatic process?

When an ideal gas is compressed isothermally,its pressure:

Vedclass Test Series

Exam Paper Generator

Online Exam Module