One mole of an ideal gas at 350 , K is in a 2 | vedclass

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

Question

(C) For an isothermal process,the heat exchanged with the surroundings is $q = -w = P_{ext} \Delta V$. Given $P_{ext} = 4 \, atm$ and $\Delta V = (3.0

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) For an isothermal process,the heat exchanged with the surroundings is $q = -w = P_{ext} \Delta V$. Given $P_{ext} = 4 \, atm$ and $\Delta V = (3.0 \, L - 2.0 \, L) = 1.0 \, L$. $q = 4 \, atm 	imes 1.0 \, L = 4 \, L \, atm$. Using the conversion $1 \, L \, atm = 101.3 \, J$,we get $q = 4 	imes 101.3 = 405.2 \, J$. The entropy change of the surroundings is given by $\Delta S_{surr} = -\frac{q_{sys}}{T}$. Since the process is isothermal and the system absorbs heat from the surroundings,$q_{surr} = -q_{sys} = -405.2 \, J$. $\Delta S_{surr} = \frac{q_{surr}}{T} = \frac{-405.2 \, J}{350 \, K} \approx -1.158 \, J \, K^{-1}$. The nearest integer is $-1$.

Similar Questions

Mixing of non-reacting gases is generally accompanied by

For which of the following processes is the entropy change $(\Delta S)$ negative?
$I)$ Sublimation of dry ice
$II)$ Freezing of water
$III)$ Crystallisation of the dissolved substance
$IV)$ Burning of rocket fuel

Calculate $\Delta S_{\text{total}}$ for a certain reaction if $\Delta H = -150 \ kJ$ and $\Delta S = 32 \ JK^{-1}$ at $300 \ K$. (in $JK^{-1}$)

At $300 \ K$,for the reaction $A \rightarrow P$,the $\Delta S_{sys}$ is $5 \ J \ K^{-1} \ mol^{-1}$. What is the heat absorbed (in $kJ \ mol^{-1}$) by the system?

Entropy is a measure of .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module