Calculate q when 0.75 mol of an ideal gas ex | vedclass

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

Question

(A) For an isothermal and reversible expansion of an ideal gas,the heat absorbed $(q)$ is given by the formula:$q = -w = 2.303 \ nRT \ \log \left( \fr

Vedclass · Accepted Answer

$955.7$

Vedclass · Answer

(A) For an isothermal and reversible expansion of an ideal gas,the heat absorbed $(q)$ is given by the formula:$q = -w = 2.303 \ nRT \ \log \left( \frac{V_2}{V_1} ight)$Given:$n = 0.75 \ mol$$T = 27 + 273 = 300 \ K$$R = 8.314 \ J \ K^{-1} \ mol^{-1}$$V_1 = 15 \ L, V_2 = 25 \ L$Substituting the values:$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log \left( \frac{25}{15} ight)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log \left( \frac{5}{3} ight)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes (\log 5 - \log 3)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes (0.70 - 0.48)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes 0.22$$q = 955.7 \ J$

Calculate $q$ when $0.75 \ mol$ of an ideal gas expands isothermally and reversibly at $27 \ ^oC$ from a volume of $15 \ L$ to $25 \ L$.
(Given $\log 5 = 0.70, \log 3 = 0.48$)
...... $Joule$

Similar Questions

Heat of reaction at constant volume is measured in which apparatus?

The heat of reaction at constant pressure is given by

Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be .............. $Joules$.

Fill in the blanks:
$(i)$ $1 \ \text{calorie} = \ldots \ldots \ \text{joules}$
$(ii)$ The work done by the system can be calculated by the equation .......

The energy diagram for a reaction $R \to P$ is shown below. The $\Delta H^o$ for the reaction will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate $q$ when $0.75 \ mol$ of an ideal gas expands isothermally and reversibly at $27 \ ^oC$ from a volume of $15 \ L$ to $25 \ L$.(Given $\log 5 = 0.70, \log 3 = 0.48$)...... $Joule$