The gas absorbs 100 J and is simultaneously | vedclass

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

Question

(C) Given: $q = +100 \ J$ (Heat absorbed). $P_{ext} = 1.50 \ atm$. $V_1 = 8 \ L, V_2 = 2 \ L$. $\Delta V = V_2 - V_1 = 2 - 8 = -6 \ L$. Work done $(w)

Vedclass · Accepted Answer

$1011$

Vedclass · Answer

(C) Given: $q = +100 \ J$ (Heat absorbed). $P_{ext} = 1.50 \ atm$. $V_1 = 8 \ L, V_2 = 2 \ L$. $\Delta V = V_2 - V_1 = 2 - 8 = -6 \ L$. Work done $(w)$ $= -P_{ext} \Delta V = -(1.50 \ atm) 	imes (-6 \ L) = 9.0 \ L \cdot atm$. Since $1 \ L \cdot atm = 101.325 \ J$,$w = 9.0 	imes 101.325 \ J = 911.925 \ J$. According to the first law of thermodynamics: $\Delta U = q + w$. $\Delta U = 100 \ J + 911.925 \ J = 1011.925 \ J \approx 1011 \ J$.

The gas absorbs $100 \ J$ and is simultaneously compressed by a constant external pressure of $1.50 \ atm$ from $8 \ L$ to $2 \ L$ volume. Hence $\Delta U$ will be ..... $J$.

Similar Questions

Which of the following relations is correct for the reaction $S(s) + O_2(g) \rightarrow SO_2(g)$?

Under isothermal condition,a gas expands from $0.2 \ dm^3$ to $0.8 \ dm^3$ against a constant external pressure of $2 \ bar$ at $300 \ K$. Find the work done by the gas. (in $J$)

Two moles of an ideal gas are allowed to expand from a volume of $10 \ dm^3$ to $2 \ m^3$ at $300 \ K$ against a pressure of $101.325 \ kPa$. Calculate the work done. (in $kJ$)

Which of the following is the correct expression of the first law of thermodynamics for an isothermal process?

For the reaction $Ag_2O_{(s)} \rightarrow 2Ag_{(s)} + \frac{1}{2}O_{2(g)}$,which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module