Five moles of hydrogen initially at STP is co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The change in internal energy $\Delta U$ for an ideal gas in an adiabatic process is given by the formula:$\Delta U = n C_v \Delta T$Since $C_v =

Vedclass · Accepted Answer

$41.50$

Vedclass · Answer

(C) The change in internal energy $\Delta U$ for an ideal gas in an adiabatic process is given by the formula:$\Delta U = n C_v \Delta T$Since $C_v = \frac{R}{\gamma - 1}$, we have:$\Delta U = n \frac{R}{\gamma - 1} (T_2 - T_1)$Given values:$n = 5 \, mol$$T_1 = 273 \, K$ (at $STP$)$T_2 = 673 \, K$$R = 8.3 \, J/mol-K$$\gamma = 1.4$Substituting these values into the equation:$\Delta U = 5 	imes \frac{8.3}{1.4 - 1} 	imes (673 - 273)$$\Delta U = 5 	imes \frac{8.3}{0.4} 	imes 400$$\Delta U = 5 	imes 8.3 	imes 1000$$\Delta U = 41500 \, J$Converting to kilo joules:$\Delta U = 41.50 \, kJ$

Five moles of hydrogen initially at $STP$ is compressed adiabatically so that its temperature becomes $673 \, K$. The increase in internal energy of the gas, in kilo joule is $(R=8.3 \, J/mol-K; \gamma=1.4$ for diatomic gas$)$

Similar Questions

Which of the following statements is not correct?

$138 \ g$ of ethyl alcohol is mixed with $72 \ g$ of water. The ratio of mole fraction of alcohol to water is

$\int_0^{\pi / 2} \sin ^8 x \cos ^2 x \, dx$ is equal to

In order to eliminate the first degree terms from the equation $2x^2 + 4xy + 5y^2 - 4x - 22y + 7 = 0$,the point to which the origin is to be shifted is:

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule,$\int_0^2 f(x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module