If standard molar enthalpy change and standard molar internal energy change measured in a bomb calorimeter are equal,which one of the following statements is correct?
- A$ \Delta n > 0 $,with increase in pressure
- B$ \Delta n > 0 $,with decrease in pressure
- C$ \Delta n < 0 $,with increase in pressure
- D$ \Delta n = 0 $,at constant pressure
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At $300 \, K$ temperature,the standard heat of formation of benzoic acid $(s)$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. At constant pressure,the heat of combustion for benzoic acid will be $..... \, kJ$.
Medium
View SolutionIdentify the correct statements from the following.
$I$. At $0 \ K$,the entropy of pure crystalline materials approaches zero.
$II$. Entropy for the process,$H_2O_{(l)} \longrightarrow H_2O_{(g)}$ decreases.
$III$. Gibbs' energy is a state function.
$I$. At $0 \ K$,the entropy of pure crystalline materials approaches zero.
$II$. Entropy for the process,$H_2O_{(l)} \longrightarrow H_2O_{(g)}$ decreases.
$III$. Gibbs' energy is a state function.
Match the following:
$A$. Isothermal process $i$. $q = \Delta U$ $B$. Adiabatic process $ii$. $W = - P \times \Delta V$ $C$. Isobaric process $iii$. $W = \Delta U$ $D$. Isochoric process $iv$. $W = - nRT \ln \left(\frac{v_f}{v_i}\right)$
| $A$. Isothermal process | $i$. $q = \Delta U$ |
| $B$. Adiabatic process | $ii$. $W = - P \times \Delta V$ |
| $C$. Isobaric process | $iii$. $W = \Delta U$ |
| $D$. Isochoric process | $iv$. $W = - nRT \ln \left(\frac{v_f}{v_i}\right)$ |
Consider the $P-V$ (pressure-volume) diagram given below,where an ideal gas is reversibly converted from state $X$ to state $Y$. Among the following,which is the correct $T-S$ (temperature-entropy) diagram corresponding to this process?
DifficultKVPY 2021
View SolutionAt $300 \ K$, $3.0 \ \text{moles}$ of an ideal gas at $3.0 \ \text{atm}$ pressure is compressed isothermally to one half of its volume by an external pressure of $6.0 \ \text{atm}$. The work done (in $kJ$) is. Given, $R=0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$ $(1 \ \text{L atm} = 101.3 \ \text{J})$.
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