Skip to main content
Back to THERMODYNAMICS

JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard
1 of 1
Application +4 | -1
Two moles of an ideal gas (Cv,m=52R)\left(C_{v, m}=\frac{5}{2} R\right) was compressed adiabatically against constant pressure of 2 atm,2 \mathrm{~atm}, which was initially at 350 K350 \mathrm{~K} and 1 atm.1 \mathrm{~atm}. The work done on the gas in this process is
View answer and explanation
Correct option: B
500R500 R
The correct answer is B (500 R) because the work done in an adiabatic process is equal to the change in internal energy, which can be calculated using the given Cv,m and temperature change.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →
Consider following reactions
I:  C(s)+12O2( g)−CO(g),ΔH1=a\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II: CO(g)+12O2( g)−CO2( g),ΔH2=b\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III: C(s)+CO2( g)−2CO(g);ΔH3=c\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}
Select the correct statement
When 0.18 g0.18 \mathrm{~g} of glucose was burnt in a bomb calorimeter, the temperature rose by 4 K.4 \mathrm{~K}. The heat capacity of the calorimeter system is (ΔCH∘=−2.8×106 J/mol glucose)\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right) Statement-1: CsI{CsI} is less soluble in water compared to CsCl{CsCl}
Satement-2: More amount of heat is liberated during the formation of CsCl{CsCl} than the formation CsI{Cs} I from their elements
Consider the following processes :

ΔH(KJ/mol)1/2A→B+1503B→2C+D−125E+A→2D+350\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

For B+D→E+2C,ΔH{B}+{D} \rightarrow {E}+2 {C}, \Delta {H} will be :
The difference b/w ΔH\Delta H and Δu\Delta u at constant volume is equal to The heat evolved in the combustion of methane is given by the following equations CH4(g)+2O2(g)→CO2(g)+H2O(l);ΔH=−890.3kJ\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ} How many grams of methane would be required to produce 44.15 kJ of heat of combustion?
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.