(C) Since the system absorbs heat,$q = +208\, J$. For a reversible isothermal expansion,the work done $w$ is given by the formula: $w = -nRT \ln(\frac

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(C) Since the system absorbs heat,$q = +208\, J$. For a reversible isothermal expansion,the work done $w$ is given by the formula: $w = -nRT \ln(\frac{V_2}{V_1})$. Given: $n = 0.04\, mol$,$R = 8.314\, J / mol\, K$,$T = 37 + 273 = 310\, K$,$V_1 = 50.0\, mL$,$V_2 = 375\, mL$. Ratio $\frac{V_2}{V_1} = \frac{375}{50} = 7.5$. Substituting the values: $w = -0.04 \times 8.314 \times 310 \times \ln(7.5)$. $w = -0.04 \times 8.314 \times 310 \times 2.01 = -207.22\, J \approx -208\, J$. Thus,$q = +208\, J$ and $w = -208\, J$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

Medium

$A$ piston filled with $0.04\, mol$ of an ideal gas expands reversibly from $50.0\, mL$ to $375\, mL$ at a constant temperature of $37\, ^oC$. It absorbs $208\, J$ of heat in this process. The value of $q$ and $w$ for the process will be: $(R = 8.314\, J / mol\, K)$ $(\ln 7.5 = 2.01)$

A
$q = -208\, J, w = +208\, J$
B
$q = +208\, J, w = +208\, J$
C
$q = +208\, J, w = -208\, J$
D
$q = -208\, J, w = -208\, J$

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First law of thermodynamics

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$A$ piston filled with $0.04\, mol$ of an ideal gas expands reversibly from $50.0\, mL$ to $375\, mL$ at a constant temperature of $37\, ^oC$. It absorbs $208\, J$ of heat in this process. The value of $q$ and $w$ for the process will be: $(R = 8.314\, J / mol\, K)$ $(\ln 7.5 = 2.01)$

Similar Questions

If $\Delta E$ is the heat of reaction for $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$ at constant volume,the $\Delta H$ (heat of reaction at constant pressure) at constant temperature is

When $1 \ mole$ of gas is heated at constant volume,the temperature rises from $273 \ K$ to $546 \ K$. If heat supplied to the gas is $x \ J$,then find the correct statement from the following.

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