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Vaishnavi Yelne
เข้าร่วมเมื่อ 21 ต.ค. 2023
#Radhe Radhe ❤️
2 kJ
of heat is released by the system and 6 kJ
of work is done on the system
#chemistry
#solved
#numerical
#4.8
#logtable
#5minutes
#easyway
#@Vaishnavi-9226
#Chemical thermodynamics
#Problem 4.8 : In a particular reaction 2 kJ
of heat is released by the system and 6 kJ
of work is done on the system. Determine
of ∆H and ∆U?
Solution : According to the first law of
thermodynamics
∆U = Q + W
Q = -2 kJ, W = +6 kJ
∆U = -2 kJ + 6 kJ = + 4 kJ
Qp
= ∆H = - 2kJ
#solved
#numerical
#4.8
#logtable
#5minutes
#easyway
#@Vaishnavi-9226
#Chemical thermodynamics
#Problem 4.8 : In a particular reaction 2 kJ
of heat is released by the system and 6 kJ
of work is done on the system. Determine
of ∆H and ∆U?
Solution : According to the first law of
thermodynamics
∆U = Q + W
Q = -2 kJ, W = +6 kJ
∆U = -2 kJ + 6 kJ = + 4 kJ
Qp
= ∆H = - 2kJ
มุมมอง: 17
วีดีโอ
Calculate standard enthalpy of reaction Given that ∆f H0 (CO2 )= -393.5 kJ mol-1, ∆f H0 (H2 O)=
มุมมอง 3317 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.10 Problem 4.10 Calculate standard enthalpy of reaction, 2C2 H6 (g) 7O2 (g) 4 CO2 (g) 6 H2 O(l) Given that ∆f H0 (CO2 )= -393.5 kJ mol-1, ∆f H0 (H2 O)= -285.8 kJ mol-1 and ∆f H0 (C2 H6 ) = -84.9 kJ mol-1 Solution - ∆r H0 = ∑ ∆f H0 (products) - ∑ ∆f H0 (reactants) = [4 ∆f H0 (CO2 ) 6 ∆f H...
Calculate the work done in oxidation of 4 moles of SO2 at 250 C if 2 SO2 (g) + O2 (g) 2 SO3 25°c
มุมมอง 3617 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.9 Problem 4.9 : Calculate the work done in oxidation of 4 moles of SO2 at 250 C if 2 SO2 (g) O2 (g) 2 SO3 (g) R = 8.314 J K-1mol-1 State whether work is done on the system or by the system. Solution : For oxidation of 4 moles of SO2 , the reaction is 4 SO2 (g) 2 O2 (g) 4 SO3 (g) W = -∆n ...
∆H for the reaction, 2C(s) + 3H2 (g) C2 H6 (g) is -84.4 kJ at 25 0 C. Calculate ∆U for the reaction
มุมมอง 7417 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.7 Problem 4.7 : ∆H for the reaction, 2C(s) 3H2 (g) C2 H6 (g) is -84.4 kJ at 25 0 C. Calculate ∆U for the reaction at 25 0 C. (R = 8.314 J K-1 mol-1) Solution : ∆H = ∆U ∆n g RT ∆n g = (moles of product gases) - (moles of reactant gases) ∆n g = 1 - 3 = -2 mol ∆H = -84.4 kJ, R = 8.314 J K-1...
300 mmol of an ideal gas occupies 13.7 dm3 at 300 K. Calculate the work done when the gas is volu
มุมมอง 2617 วันที่ผ่านมา
12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.6 #Problem 4.6 : 300 mmol of an ideal gas occupies 13.7 dm3 at 300 K. Calculate the work done when the gas is expanded until its volume has increased by 2.3 dm3 (a) isothermally against a constant external pressure of 0.3 bar (b) isothermally and reversibly (c) into vacuum. Solution : a. ...
22 g of CO2 are compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa
มุมมอง 2617 วันที่ผ่านมา
Problem 4.5 : 22 g of CO2 are compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa when the work obtained is 1.2 kJ. Find the final pressure. Solution : W = -2.303 nRT log10 P1 P2 n = 22 g 44 g mol-1 =0.5 mol, T = 298 K, P1 = 100 kPa, W = 1.2 kJ = 1200 J Hence, 1200 J = -2.303 × 0.5 mol ×8.314 J K-1 mol-1 × 298K × log10 100 kPa P2 or log10 100 kPa P2 = -1200 2.303 ×0...
2 moles of an ideal gas are expanded isothermally and reversibly from 20 L to 30 L at 300 K.
มุมมอง 9417 วันที่ผ่านมา
Problem 4.4 : 2 moles of an ideal gas are expanded isothermally and reversibly from 20 L to 30 L at 300 K. Calculate the work done (R= 8.314 J K-1 mol-1) Solution : Wmax = -2.303 nRT log10 V2 V1 n = 2 mol, T = 300 K, V1 = 20 L, V2 = 30 L, R = 8.314 J/K mol Substitution of these quantities into the equation gives Wmax = -2.303 × 2 mol × 8.314 J/K mol × 300K × log10 30 L 20 L = -2.303 × 2 × 8.314...
Calculate the constant external pressure required to compress 2 moles of an ideal gas from vol
มุมมอง 3917 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.2 Problem 4.2 : Calculate the constant external pressure required to compress 2 moles of an ideal gas from volume of 25 dm3 to 13 dm3 when the work obtained is 4862.4 J. Solution : W = - Pext ∆V = - Pext (V2 - V1 ) V1 = 25 dm3 , V2 = 13 dm3 , W = 4862.4 J W = 4862.4 J × dm3 bar 100 J = 4...
Three moles of an ideal gas are expanded isothermally from 15 dm3 to 20 dm3 at constant external pre
มุมมอง 2617 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #chemical thermodynamics #@Vaishnavi-9226 #4.1 # Three moles of an ideal gas are expanded isothermally from 15 dm3 to 20 dm3 at constant external pressure of 1.2 bar. Estimate the amount of work in dm3 bar and J. Solution : W = - Pext ∆V = - Pext (V2 - V1 ) Pext = 1.2 bar, V1 = 15 dm3 , V2 = 20 dm3 Substitution of these quantities ...
If 20.0 cm3 of 0.050 M Ba(NO3 )2 are mixed with 20.0 cm3 of 0.020M NaF, will BaF2 precipitate?
มุมมอง 1917 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #Ionic equilibria #@Vaishnavi-9226 #3.13 Problem 3.13 : If 20.0 cm3 of 0.050 M Ba(NO3 )2 are mixed with 20.0 cm3 of 0.020 M NaF, will BaF2 precipitate ? K sp of BaF2 is 1.7 × 10-6 at 298 K. Solution : Final volume of solution is 20 20 = 40 cm3 , [Ba(NO3 )2 ] = 0.050 × 20 40 = 0.025 M [NaF] = 0.020 × 20 40 = 0.010M
The solubility product of AgBr is 5.2 × 10-13. Calculate its solubility in mol dm-3 and g dm-3(Mol
มุมมอง 6022 วันที่ผ่านมา
Problem 3.12 : The solubility product of AgBr is 5.2 × 10-13. Calculate its solubility in mol dm-3 and g dm-3(Molar mass of AgBr = 187.8 g mol-1) Solution : The solubility equilibrium of AgBr is : AgBr(s) Ag⊕(aq) Br (aq) x = 1, y = 1 K sp = [Ag⊕][Br ] = S2 S = K sp = 5.2 × 10-13 = 7.2 × 10-7 mol dm-3 The solubility in g dm-3 = molar solubility in mol dm-3 × molar mass g mol-1 S = 7.2 × 10-7 mol...
A solution is prepared by mixing equal volumes of 0.1M MgCl2 and 0.3M Na2 C2O4 at 293k.
มุมมอง 4625 วันที่ผ่านมา
#12th #chemistry #solved numerical #5minutes #easyway #solution #Ionic equilibria #@Vaishnavi-9226 #3.11 #A solution is prepared by mixing equal volumes of 0.1M MgCl2 and 0.3M Na2 C2 O4 at 293 K. Would MgC2 O4 precipitate out ? K sp of MgC2 O4 at 293 K is 8.56 × 10-5. Thank you for visiting
Calculate the pH of buffer solution composed of 0.1Mweak base BOH and 0.2m of its salt BA.
มุมมอง 12429 วันที่ผ่านมา
#Calculate the pH of buffer solution composed of 0.1Mweak base BOH and 0.2m of its salt BA. 12th #chemistry #ionic equilibrium #3.10 #log table #5minutes #solved numerical #@Vaishnavi-9226
Calculate PH of buffer solution containing 0.05 mol NaF per litre and 0.015 mol HF per litre.
มุมมอง 56หลายเดือนก่อน
#Calculate pH of buffer solution containing 0.05 mol NaF per litre and 0.015 mol HF per litre. #12th #chemistry #ionic equilibrium #3.9 #log table #5minutes #solved numerical #@Vaishnavi-9226 Thank you for visiting
App bhi 12th mai hai kyaa😅
@@Itx..sanket-07 Nahi
Yes I will
Pls. Next chp ka bhi video dalo ma'am
Mam please upload more questions l'm completely dependent on you
@@AnjumKhan-oi5yt sorry for the delay
Upload more questions of chp 4 and 5 please
Mam please upload the next questions
Very helpful 👍🙏
Awesome lecture mam
😊😊😊
Wonderful 🎉
You are the only one who guve explanation of the calculations also😊
Thank you for watching! ❤️
You are the best teacher ❤❤
❤❤❤
Thank you so much mam
Very wonderful madam 🎉
Thankyou mam ❤❤
You're welcome! 😊
❤❤
very helpful
Thank you for watching! 😄
Superb
Didi aaj nhi kiya aapne upload ❤️
Superb mam ! Just because of you mam if God wills I will score full marks in physical chemistry 😊,please continue this series till chp 6
@@AnjumKhan-oi5ytThank you so much ❤️ Yes I will
❤❤❤
Keep going mam❤
Didi aap bhot acche se samjate ho me baki channel par dekhti hun ki sirf given btate hai aur chod dete hai calculation hi nhi batate thank you so much ☺️❤ didi please pyq bhi solve krake do ❤😊
Didi please jaldi jaldi upload kro na exam bhot paas me hai
Sorry for delaying videos.Because My exams are also so near 😢😢
Didi please electrochemistry ke numericals btao na ❤😊
@swejalchakole1373 yes I will
Good brother thanks
❤😊
Thanks mam your efforts means alot for us 😊
@@anjumkhan9354 Thank you so much for watching my videos ♥️👍🏻
Mam aap aise hi continue kariye ek din aapke videos per bahot saare views aaenge Kyunki aapke explanation bahot achhi hai aap calculation bhi explain karte ho jo TH-cam per koi nahi karta
@@AnjumKhan-oi5yt thank you so much ❤️
Thank you so much
Mam aapke explanation ki wajah se mera numerical ekdam achhe se horaha hai exam me solve karke aaungi
@@AnjumKhan-oi5yt Nice ☺️
Mam please aap mujhe explain karsakte hain kya ki yahan per log1-log 308.9 ko Inka log aaya 0.0000-2.4899 phir aapne isko subtract kaise kiye aur baaki sab ekdam acche se samajh Aya.
I will definitely explain it.
0000-4899 ko simply subtract Karo then 0.0000-2.4899 Mai -2mai 1add hota hai to vo -3ban jata hai ans milta hai -3.5101.Ab try karke dekho aata hai ky ... otherwise you can also ask me again
Please aise hi CHP 1 ke saare solve unsolved questions karaiye
@@anjumkhan9354 All unsolved questions of 1lesson are posted on channel 😊
Mam aapka explanation bahot achha hai mai ye sum jab bhi karne baithti thi calculation per aakar stuck hojati thi . thank you so much
Didi please take daily lecture
Didi please physics also ❤😊
@@swejalchakole1373 Yes .. Just after completed all the chemistry numericals.
@Vaishnavi-9226 thank you so much didi 💕👌
👍🏻 5:19
Didi please physics ke bhi kro na 😊❤️ aap pure details me pdhate ho bohot accha kar rhe ho didi 💕👌💯
Thank you❤
Didi 1.2vala upload kro 😊
Mam please take daily lecture ❤😊
@@swejalchakole1373 Yes I will
@@swejalchakole1373 Are you from which class ?
Bhot accha samj aat hIi❤
Didi please physics numerical ki krao na
Nice👌
❤❤❤❤❤❤❤❤❤😊
Wow mam aap bhot aache se samjate ho
Super mam
Formula mein values put karne ke baad unka log lena hi hota hai agar direct multiply kare to answer nhi ata hai kya
Thoda hard jata hai ..log say humey calculate nahi karna padta hai bass log dek kar add or subtract karna padta
Haan kar sakte direct calculate bhi.
👍
👍🏻