(B) According to the first law of thermodynamics,$\Delta U = q + w$. Given: Heat absorbed,$q = +84\, kJ$. Work done by the system,$w = -P \cdot \Delta

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(B) According to the first law of thermodynamics,$\Delta U = q + w$. Given: Heat absorbed,$q = +84\, kJ$. Work done by the system,$w = -P \cdot \Delta V$. Pressure,$P = 1\, MPa = 10^6\, Pa$. Change in volume,$\Delta V = 0.06\, m^3 - 0.03\, m^3 = 0.03\, m^3$. Calculating work done: $w = - (10^6\, Pa) \times (0.03\, m^3) = -30,000\, J = -30\, kJ$. Calculating change in internal energy: $\Delta U = 84\, kJ + (-30\, kJ) = 54\, kJ$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

Medium

$A$ mixture of gas expands from $0.03\, m^3$ to $0.06\, m^3$ at a constant pressure of $1\, MPa$ and absorbs $84\, kJ$ of heat during the process. The change in internal energy of the mixture is $.....\, kJ$.

A
$30$
B
$54$
C
$84$
D
$3132$

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Thermodynamics

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First law of thermodynamics

The enthalpy of vaporization of water at $100 \, ^\circ C$ is $40.63 \, kJ \, mol^{-1}$. The value of $\Delta U$ for this process is ....... $kJ \, mol^{-1}$.

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For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$ at $298 \ K$,the enthalpy change $(\Delta H)$ is $-92.38 \ kJ$. What will be the internal energy change $(\Delta U)$ at $298 \ K$ in $kJ$?

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$A$ gas is allowed to expand in an insulated container against a constant external pressure of $10 \ atm$ from an initial volume $4.2 \ L$ to $5.2 \ L$. The change in internal energy $\Delta E$ of the gas will be $:-$

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$5 \, mol$ of an ideal gas expands isothermally and irreversibly from a pressure of $10 \, atm$ to $1 \, atm$ against a constant external pressure of $1 \, atm$. The work done $(w_{irr})$ at $300 \, K$ is:

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$A$ mixture of gas expands from $0.03\, m^3$ to $0.06\, m^3$ at a constant pressure of $1\, MPa$ and absorbs $84\, kJ$ of heat during the process. The change in internal energy of the mixture is $.....\, kJ$.

Similar Questions

The enthalpy of vaporization of water at $100 \, ^\circ C$ is $40.63 \, kJ \, mol^{-1}$. The value of $\Delta U$ for this process is ....... $kJ \, mol^{-1}$.

$A$ certain mass of an ideal gas absorbs $80 \ kJ$ heat and the gas is expanded from $2 \ L$ to $10 \ L$ at a constant pressure of $25 \ bar$. What is $\Delta U$ for the gas in the process in $kJ$? $(1 \ bar-L = 100 \ J)$

For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$ at $298 \ K$,the enthalpy change $(\Delta H)$ is $-92.38 \ kJ$. What will be the internal energy change $(\Delta U)$ at $298 \ K$ in $kJ$?

$A$ gas is allowed to expand in an insulated container against a constant external pressure of $10 \ atm$ from an initial volume $4.2 \ L$ to $5.2 \ L$. The change in internal energy $\Delta E$ of the gas will be $:-$

$5 \, mol$ of an ideal gas expands isothermally and irreversibly from a pressure of $10 \, atm$ to $1 \, atm$ against a constant external pressure of $1 \, atm$. The work done $(w_{irr})$ at $300 \, K$ is:

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